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Title: Birkhauser Inequalities A Mathematical Olympiad Approach Oct.2009
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Radmila Bulajich Manfrino
José Antonio Gómez Ortega
Rogelio Valdez Delgado

Inequalities
A Mathematical Olympiad Approach

Birkhäuser
Basel · Boston · Berlin

Autors:
Radmila Bulajich Manfrino
Rogelio Valdez Delgado
Facultad de Ciencias
Universidad Autónoma Estado de Morelos
Av
...
Chamilpa
62209 Cuernavaca, Morelos
México
e-mail: bulajich@uaem
...
mx

José Antonio Gómez Ortega
Departamento de Matemàticas
Facultad de Ciencias, UNAM
Universidad Nacional Autónoma de México
Ciudad Universitaria
04510 México, D
...

México
e-mail: jago@fciencias
...
mx

2000 Mathematical Subject Classiﬁcation 00A07; 26Dxx, 51M16
Library of Congress Control Number: 2009929571
Bibliograﬁsche Information der Deutschen Bibliothek
Die Deutsche Bibliothek verzeichnet diese Publikation in der Deutschen Nationalbibliograﬁe; detaillierte bibliograﬁsche Daten sind im Internet über ...
de>
abrufbar
...
All rights are reserved, whether the whole or part of the
material is concerned, speciﬁcally the rights of translation, reprinting, re-use of illustrations, recitation, broadcasting, reproduction on microﬁlms or in other ways, and storage
in data banks
...

© 2009 Birkhäuser Verlag AG
Basel · Boston · Berlin
Postfach 133, CH-4010 Basel, Schweiz
Ein Unternehmen von Springer Science+Business Media
Gedruckt auf säurefreiem Papier, hergestellt aus chlorfrei gebleichtem Zellstoff
...
birkhauser
...
In this volume we present both classic inequalities
and the more useful inequalities for confronting and solving optimization problems
...

The book has been organized in four chapters which have each of them a
diﬀerent character
...
Most of
them are numerical inequalities generally lacking any geometric meaning
...
We emphasize the importance of some of these inequalities, such as
the inequality between the arithmetic mean and the geometric mean, the CauchySchwarz inequality, the rearrangement inequality, the Jensen inequality, the Muirhead theorem, among others
...
We also emphasize how the substitution strategy is used to deduce several
inequalities
...
There we apply basic numerical inequalities, as described in Chapter 1, to geometric problems
to provide examples of how they are used
...
We introduce examples where the symmetrical properties of the variables help to solve some problems
...
We also include several classic geometric
problems, indicating the methods used to solve them
...

vi

Introduction

In Chapter 4 we provide solutions to each of the two hundred and ten exercises in Chapters 1 and 2, and to the problems presented in Chapter 3
...
This is why we do not give individual credits for them
...
This indicates that the technique outlined in the corresponding section
can be used as a tool for solving the particular exercise
...
These students were developing their skills and mathematical knowledge in preparation for the international
competitions in which Mexico participates
...

Contents
Introduction
1 Numerical Inequalities
1
...

1
...
3 A fundamental inequality,
arithmetic mean-geometric mean
...
4 A wonderful inequality:
The rearrangement inequality
...
5 Convex functions
...
6 A helpful inequality
...
7 The substitution strategy
...
8 Muirhead’s theorem
...

...

7

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

13
20
33
39
43

2 Geometric Inequalities
2
...

2
...

2
...
4 Euler’s inequality and some applications
...
5 Symmetric functions of a, b and c
...
6 Inequalities with areas and perimeters
...
7 Erd˝s-Mordell Theorem
...
8 Optimization problems
...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

101

4 Solutions to Exercises and Problems
117
4
...
117
4
...
140
4
...
162
Notation

205

viii

Contents
Bibliography

207

Index

209

Chapter 1

Numerical Inequalities
1
...
The
order of the real numbers enables us to compare two numbers and to decide which
one of them is greater or whether they are equal
...
We will also assume the
following three properties
...
1
...
Every real number x has one and only one of the following properties:
(i) x = 0,
(ii) x ∈ P (that is, x > 0),
(iii) −x ∈ P (that is, −x > 0)
...
1
...
If x, y ∈ P , then x+y ∈ P (in symbols x > 0, y > 0 ⇒ x+y > 0)
...
1
...
If x, y ∈ P , then xy ∈ P (in symbols x > 0, y > 0 ⇒ xy > 0)
...
In general the number one is set on the right hand
side of 0
...
1
...

Now we can deﬁne the relation a is greater than b if a − b ∈ P (in symbols
a > b)
...
Observe that

2

Numerical Inequalities

a a
...

We will denote by R the set of real numbers and by R+ the set P of positive
real numbers
...
1
...

(i) If a 0 ⇔
b − a > 0 ⇔ a b or a 0
...

(iii) a < b, b < c ⇒ a < c
...

(v) a 0, we have that

a
b

> 1 if and only if a > b
...

Geometrically, |x| is the distance of the number x (on the real line) from the origin
0
...

1
...
4
...

(i) |x| ≥ 0, and is equal to zero only when x = 0
...

2

(iii) |x| = x2
...

(v)

a
|a|
=
, with b = 0
...
1
...
The triangle inequality states that for any
pair of real numbers a and b,
|a + b| ≤ |a| + |b|
...

Proof
...
3 it is suﬃcient to verify that |a + b|2 ≤ (|a| + |b|)2 :
2

2

2

2

2

|a + b| = (a + b)2 = a2 + 2ab + b2 = |a| + 2ab + |b| ≤ |a| + 2 |ab| + |b|
2

2

2

= |a| + 2 |a| |b| + |b| = (|a| + |b|)
...
Note that, when ab ≥ 0, we can deduce that ab = |ab| = |a| |b|, and then
the equality holds
...
, xn ,
is
|x1 + x2 + · · · + xn | ≤ |x1 | + |x2 | + · · · + |xn |
...
This can be proved in a similar
way or by the use of induction
...

Exercise 1
...
Let x, y, a, b be real numbers, prove that
(i) |x| ≤ b ⇔ −b ≤ x ≤ b,
(ii) ||a| − |b|| ≤ |a − b|,
(iii) x2 + xy + y 2 ≥ 0,
(iv) x > 0, y > 0 ⇒ x2 − xy + y 2 > 0
...
6
...

4

Numerical Inequalities

Exercise 1
...
Let a, b be real numbers such that 0 ≤ a ≤ b ≤ 1
...

4
(i) 0 ≤

Exercise 1
...
Prove that if n, m are positive integers, then
√
2 < m+2n
...
9
...

√
2
√
2
Exercise 1
...
If x, y > 0, then x + y ≥ x + y
...
11
...

Exercise 1
...
Let f (a, b, c, d) = (a − b)2 + (b − c)2 + (c − d)2 + (d − a)2
...

Exercise 1
...
(IMO, 1960) For which real values of x the following inequality
holds:
4x2
√
< 2x + 9?
(1 − 1 + 2x)2
√
Exercise 1
...
Prove that for any positive integer n, the fractional part of 4n2 + n
is smaller than 1
...
15
...
Prove that
ab
bc
ca
+
+
≤ 1
...
2 The quadratic function ax2 + 2bx + c
One very useful inequality for the real numbers is x2 ≥ 0, which is valid for any
real number x (it is suﬃcient to consider properties 1
...
1, 1
...
3 and Exercise 1
...
The use of this inequality leads to deducing many other
inequalities
...
These quadratic functions appear frequently in
optimization problems or in inequalities
...
2 The quadratic function ax2 + 2bx + c

5

One common example consists in proving that if a > 0, the quadratic function
2
b
ax2 + 2bx + c will have its minimum at x = − a and the minimum value is c − ba
...

a

2

b
Since x + a ≥ 0 and the minimum value of this expression, zero, is attained
b
when x = − a , we conclude that the minimum value of the quadratic function is
2
c − ba
...
In fact, since ax2 +2bx+c = a x +
a
2

2
b 2
+c− ba
a

b
and since a x + a ≤ 0 (because a < 0), the greatest value of this last expression
2
is zero, thus the quadratic function is always less than or equal to c − ba and
b
assumes this value at the point x = − a
...
2
...
If x, y are positive numbers with x + y = 2a, then the product xy
is maximal when x = y = a
...
Hence, xy = x(2a − x) = −x2 + 2ax =
−(x − a)2 + a2 has a maximum value when x = a, and then y = x = a
...
In fact, if x, y are the
lengths of the sides of the rectangle, the perimeter is 2(x + y) = 4a, and its area
is xy, which is maximized when x = y = a
...
2
...
If x, y are positive numbers with xy = 1, the sum x + y is minimal
when x = y = 1
...
Therefore,
x = y = 1
...

This can also be interpreted geometrically in the following way, “of all the
rectangles with area 1, the square has the smallest perimeter”
...
Moreover, the perimeter is 4 if
√
1
and only if x − √x = 0, that is, when x = y = 1
...
2
...
For any positive number x, we have x +

1
x

≥ 2
...
Moreover, the equality holds if
Observe that x + x =
√
1
and only if x − √x = 0, that is, when x = 1
...
2
...
If a, b > 0, then
a = b
...

b
Example 1
...
5
...

Remember that a, b and c are the lengths of the sides of a triangle if and
only if a + b > c, a + c > b and b + c > a
...

Now, [b + c − a][c + a − b][a + b − c] > 0 if the three factors are positive or if one of
them is positive and the other two are negative
...
Therefore the three factors are necessarily
positive
...
16
...
Prove that the roots are
real and that they belong to the interval −1 ≤ x ≤ 1
...
17
...

4
4
4
1 A quadratic function ax2 + bx + c with a > 0 is positive when its discriminant Δ = b2 − 4ac
2
b
is negative, in fact, this follows from ax2 + bx + c = a(x + 2a )2 + 4ac−b
...

then ax

1
...
3 A fundamental inequality,
arithmetic mean-geometric mean
The ﬁrst inequality that we consider, fundamental in optimization problems, is
the inequality between the arithmetic mean and the geometric mean of two nonnegative numbers a and b, which is expressed as
a+b √
≥ ab,
(AM-GM)
...

if
The numbers a+b and ab are known as the arithmetic mean and the ge2
ometric mean of a and b, respectively
...

2
2
2
√
√
And the equality holds if and only if a = b, that is, when a = b
...
18
...

Exercise 1
...
For x > 0, prove that x +

1
x

+

≥ 2
...
20
...

Exercise 1
...
For x, y ∈ R+ , prove that 2(x2 + y 2 ) ≥ (x + y)2
...
22
...
23
...
24
...
25
...

2
1 (a−b)
8
a

≤

a+b
2

4
x+y
...

√
ab ≤

2
1 (a−b)

...

(1
...
Let A be the point where the perpendicular to BC in D intersects the
semicircle and let E be the perpendicular projection from D to the radius AO
...
Since ABD and CAD are similar right triangles,
we deduce that
h
x
√
= ,
then
h = xy
...

then
g=
√ = x+y ,
1
1
xy
x+y
2
x + y
Finally, the geometry tells us that in a right triangle, the length of one leg is
always smaller than the length of the hypotenuse
...

xy ≤
1
1 ≤
2
+y
x
The number

2

1
1
x+y

is known as the harmonic mean of x and y, and the left inequality

in (1
...

Some inequalities can be proved through the multiple application of a simple
inequality and the use of a good idea to separate the problem into parts that are
easier to deal with, a method which is often used to solve the following exercises
...
26
...

Exercise 1
...
For x, y, z ∈ R, x2 + y 2 + z 2 ≥ xy + yz + zx
...
28
...

Exercise 1
...
For x, y ∈ R, x2 + y 2 + 1 ≥ xy + x + y
...
30
...
31
...

Exercise 1
...
For x, y, z ∈ R, x2 + y 2 + z 2 ≥ x

+

√1
yz

+

√1
...

The inequality between the arithmetic mean and the geometric mean can
be extended to more numbers
...

≥

≥

1
...
Moreover, the
equality holds if and only if a = b, c = d and ab = cd, that is, when the numbers
satisfy a = b = c = d
...
33
...

Exercise 1
...
For a, b, c, d ∈ R+ , (a + b + c + d)
Exercise 1
...
For a, b, c, d ∈ R+ ,

a
b

+

b
c

+

c
d

+

d
a

1
a

+

1
b

+

1
c

+

1
d

≥ 16
...

√
A useful trick also exists for checking that the inequality a+b+c ≥ 3 abc
3
is true for any three non-negative numbers a, b and c
...
Since the AM-GM inequality holds for four
√
4
a+b+c+d
numbers, we have
≥ 4 abcd = d3 d = d
...

4
4
4
4
√
3
Hence, a+b+c ≥ d = abc
...
If a1 , a2 ,
...

These numbers are known as the arithmetic mean and the geometric mean of the
numbers a1 , a2 ,
...

Theorem 1
...
1 (The AM-GM inequality)
...

n
First proof (Cauchy)
...
We will
proceed by mathematical induction on n, but this is an induction of the following
type
...

(2) We prove that Pn ⇒ Pn−1
...

When (1), (2) and (3) are veriﬁed, all the assertions Pn with n ≥ 2 are shown
to be true
...

(1) This has already been done in the ﬁrst part of the section
...
, an−1 be non-negative numbers and let g = n−1 a1 · · · an−1
...
e
...
, an−1 , we get n
numbers to which we apply Pn ,
a1 + · · · + an−1 + g
√
≥ n a1 a2 · · · an−1 g =
n

n

g n−1 · g = g
...

a1 +···+an−1
n−1

≥

(3) Let a1 , a2 ,
...

We have applied the statement P2 several times, and we have also applied the
√
√
√
statement Pn to the numbers a1 a2 , a3 a4 ,
...

Second proof
...
We take two numbers ai , one smaller than A
n
and the other greater than A (if they exist), say a1 = A − h and a2 = A + k, with
h, k > 0
...

Since a1 + a2 = A + A + k − h = A − h + A + k = a1 + a2 , clearly a1 + a2 + a3 +
· · · + an = a1 + a2 + a3 + · · · + an , but a1 a2 = A(A + k − h) = A2 + A(k − h) and
a1 a2 = (A + k)(A − h) = A2 + A(k − h) − hk, then a1 a2 > a1 a2 and thus it follows
that a1 a2 a3 · · · an > a1 a2 a3 · · · an
...
Since every time we perform this
operation we create a number equal to A, this process can not be used more than
n times
...
3
...
Find the maximum value of x(1 − x3 ) for 0 ≤ x ≤ 1
...
If
y = x(1 − x3 ), it is clear that the right side of 3y 3 = 3x3 (1 − x3 )(1 − x3 )(1 − x3 ),
expressed as the product of four numbers 3x3 , (1 − x3 ), (1 − x3 ) and (1 − x3 ), has
a constant sum equal to 3
...

434
1
√
...

Moreover, the maximum value is reached using 3x3 = 1 − x3 , that

1
...
36
...
, n
...

Exercise 1
...
If {a1 ,
...
, bn } ⊂ R+ , then
a1
a2
an
+
+ ···+
≥n
b1
b2
bn

b1
b2
bn
+
+ ···+
≥ n
...
38
...

Exercise 1
...
If a, b, c > 0 and (1 + a)(1 + b)(1 + c) = 8, then abc ≤ 1
...
40
...

Exercise 1
...
For non-negative real numbers a, b, c, prove that
a2 b2 + b2 c2 + c2 a2 ≥ abc(a + b + c)
...
42
...

Exercise 1
...
If a, b, c > 0 satisfy that abc = 1, prove that
1 + ab 1 + bc 1 + ac
+
+
≥ 3
...
44
...
45
...

a+b+c

1
+ · · · + n , prove that
1

n(n + 1) n < n + Hn

for n ≥ 2
...
46
...
, xn > 0 such that

1
1+x1

1
+ · · · + 1+xn = 1
...

Exercise 1
...
(Short list IMO, 1998) Let a1 , a2 ,
...

(a1 + a2 + · · · + an ) (1 − a1 ) (1 − a2 ) · · · (1 − an )
n

12

Numerical Inequalities

Exercise 1
...
Let a1 , a2 ,
...
Prove that
√

a1 + · · · +

√

an ≥ (n − 1)

1
1
√ + ··· + √
a1
an

1
1
1+a1 +· · ·+ 1+an

=

...
49
...
, an , b1 , b2 ,
...
Prove that
a2
a2
1
1
n
+ ···+
≥ (a1 + · · · + an )
...
50
...

3 + abc
3 + abc
3 + abc
+b
b +c
c +a
abc

Exercise 1
...
Let a, b, c be positive numbers with a + b + c = 1, prove that
1
+1
a

1
+1
b

1
+1
c

≥ 64
...
52
...

Exercise 1
...
(Czech and Slovak Republics, 2005) Let a, b, c be positive numbers
that satisfy abc = 1, prove that
b
c
3
a
+
+
≥
...
54
...

1
1+a

+

1
1+b

+

1
1+c

= 1
...
55
...

a+b b+c c+a
Exercise 1
...
Let a1 , a2 ,
...
, bn be positive numbers, prove that
n
i=1

1
ai b i

n

2

(ai + bi ) ≥ 4n2
...
57
...

3

1
...
58
...

Exercise 1
...
(Russia, 1992) For any real numbers x, y > 1, prove that
x2
y2
+
≥ 8
...
4 A wonderful inequality:
The rearrangement inequality
Consider two collections of real numbers in increasing order,
a1 ≤ a2 ≤ · · · ≤ a n

and b1 ≤ b2 ≤ · · · ≤ bn
...
, an ) of (a1 , a2 ,
...

(1
...
3)

Moreover, the equality in (1
...
, an ) = (a1 , a2 ,
...

And the equality in (1
...
, an ) = (an , an−1 ,
...

Inequality (1
...

Corollary 1
...
1
...
, an ) of (a1 , a2 ,
...

1
2
n
Corollary 1
...
2
...
, an ) of (a1 , a2 ,
...

a1
a2
an
Proof (of the rearrangement inequality)
...
Let
S = a1 b 1 + a2 b 2 + · · · + ar b r + · · · + as b s + · · · + an b n ,
S = a 1 b 1 + a2 b 2 + · · · + as b r + · · · + ar b s + · · · + an b n
...
Hence
S − S = ar br + as bs − as br − ar bs = (bs − br )(as − ar )
...
Repeating this process we get
the result that the sum S is maximal when a1 ≤ a2 ≤ · · · ≤ an
...
4
...
(IMO, 1975) Consider two collections of numbers x1 ≤ x2 ≤ · · · ≤
xn and y1 ≤ y2 ≤ · · · ≤ yn , and one permutation (z1 , z2 ,
...
, yn )
...

By squaring and rearranging this last inequality, we ﬁnd that it is equivalent
to

n

x2 − 2
i

i=1
n
i=1

but since
equivalent to

n

n

xi yi +
i=1

2
yi

=

n
2
i=1 zi ,

2
yi ≤

i=1

n

x2 − 2
i

i=1

n

n

xi zi +
i=1

2
zi ,

i=1

then the inequality we have to prove turns to be
n

n

xi zi ≤
i=1

xi yi ,
i=1

which in turn is inequality (1
...

Example 1
...
4
...
, xn be distinct positive integers, prove
that
x1
1
x2
xn
1 1
+ 2 + ···+ 2 ≥ + + ··· +
...
, an ) be a permutation of (x1 , x2 ,
...
, bn ) = n2 , (n−1)2 ,
...
, n
...
, an ) of (a1 , a2 ,
...
, n
...
3) we can argue that

x1
x2
xn
+ 2 + · · · + 2 = a 1 b 1 + a2 b 2 + · · · + an b n
2
1
2
n
≥ an b1 + an−1 b2 + · · · + a1 bn
= a1 bn + a2 bn−1 + · · · + an b1
a1
a2
an
= 2 + 2 + ···+ 2
...
, n ≤ an , we have that
1
x1 x2
xn
a 1 a2
an
1
2
n
1 1
+
+···+ 2 ≥ 2 + 2 +···+ 2 ≥ 2 + 2 +···+ 2 = + +···+
...
4
...
(IMO, 1964) Suppose that a, b, c are the lengths of the sides of a
triangle
...

1
...
In this case, a (b + c − a) ≤
b (a + c − b) ≤ c (a + b − c)
...

By (1
...

Therefore, 2 a2 (b + c − a) + b2 (c + a − b) + c2 (a + b − c) ≤ 6abc
...
4
...
(IMO, 1983) Let a, b and c be the lengths of the sides of a triangle
...

Consider the case c ≤ b ≤ a (the other cases are similar)
...
2) leads us to
b
c
1
1
1
a(b + c − a) + b(c + a − b) + c(a + b − c)
a
b
c
1
1
1
≥ a(b + c − a) + b(c + a − b) + c(a + b − c)
...

c
a
b
b(c−b)
a

+

c(a−c)
b

≤ 0
...

Example 1
...
7 (Cauchy-Schwarz inequality)
...
, xn , y1 ,
...

i=1

The equality holds if and only if there exists some λ ∈ R with xi = λyi for all
i = 1, 2,
...

16

Numerical Inequalities

If x1 = x2 = · · · = xn = 0 or y1 = y2 = · · · = yn = 0, the result is evident
...

yi
xi
Take ai = S and an+i = T for i = 1, 2,
...
Using Corollary 1
...
1,
n

2=
i=1

x2
i
+
S2

n
i=1

2
yi
=
T2

2n

a2
i
i=1

≥ a1 an+1 + a2 an+2 + · · · + an a2n + an+1 a1 + · · · + a2n an
x1 y1 + x2 y2 + · · · + xn yn

...
, n, or equivalently, if
S
and only if xi = T yi for i = 1, 2,
...

Another proof of the Cauchy-Schwarz inequality can be established using
Lagrange’s identity
2

n

xi yi
i=1

n

=

x2
i

i=1

n

2
yi −

i=1

1
2

n

n

(xi yj − xj yi )2
...

Example 1
...
8 (Nesbitt’s inequality)
...

b+c c+a a+b
2
Without loss of generality, we can assume that a ≤ b ≤ c, and then it follows
1
1
1
that a + b ≤ c + a ≤ b + c and b+c ≤ c+a ≤ a+b
...
2) twice, we obtain
a
b
c
b
c
a
+
+
≥
+
+
,
b+c c+a a+b
b+c c+a a+b
a
b
c
c
a
b
+
+
≥
+
+

...

Another way to prove the inequality is using Inequality (1
...

b+c
c+a a+b

1
...

b+c c+a a+b

2b+c+a
c+a

+

2c+a+b
a+b

≥ 6,

Example 1
...
9
...

Prove that
1
1
1
3
+
+
≥
...
Let x = a , y =
1
and z = c , thus

S=
=

=

1
b

1
1
1
+ 3
+ 3
+ c) b (c + a) c (a + b)

a3 (b

x3
1 1
+
y z

+

y3
1 1
+
z x

+

z3
1
1
+
x y

y2
z2
x2
+
+

...
Using the rearrangement inequality (1
...
Therefore, S ≥ 3
...
4
...
(APMO, 1998) Let a, b, c ∈ R+ , prove that
1+

a
b

1+

b
c

1+

c
a+b+c
≥2 1+ √
3
a
abc

b
c

1+

c
a+b+c
≥2 1+ √
3
a
abc

...

3
b
c a
c
b a
abc

18

Numerical Inequalities

Now we set a = x3 , b = y 3 , c = z 3
...

+ 3+ 3+ 3 + 3+ 3 ≥
y3
z
x
z
y
x
xyz
But, if we consider
(a1 , a2 , a3 , a4 , a5 , a6 ) =
(a1 , a2 , a3 , a4 , a5 , a6 ) =
(b1 , b2 , b3 , b4 , b5 , b6 ) =

x y z x z y
,
, , , , ,
y z x z y x
y z x z y x
, , , , ,
,
z x y y x z
x2 y 2 z 2 x2 z 2 y 2
, , , , ,
y 2 z 2 x2 z 2 y 2 x2

,

we are led to the following result:
x3
z 2 x x2 z
z2 y
y2 x
y3
z3
x3
z3
y3
x2 y y 2 z
+ 3+ 3+ 3 + 3+ 3 ≥ 2 + 2 + 2 + 2 + 2 + 2
y3
z
x
z
y
x
y z
z x x y
z y y x x z
=

y2
z2
x2
z2
y2
x2
+
+
+
+
+
yz zx xy zy yx xz

=

2 x3 + y 3 + z 3

...
4
...
Let a1 ≤ a2 ≤ · · · ≤ an and b1 ≤ b2 ≤
· · · ≤ bn , then
a 1 b 1 + a2 b 2 + · · · + an b n
a 1 + a2 + · · · + an b 1 + b 2 + · · · + b n
≥
·

...

...

...

...

...

a1 b1 + · · · + an bn ≥ a1 bn + a2 b1 + · · · + an bn−1 ,
and adding together all the expressions, we obtain
n (a1 b1 + · · · + an bn ) ≥ (a1 + · · · + an ) (b1 + · · · + bn )
...

Exercise 1
...
Any three positive real numbers a, b and c satisfy the following
inequality:
a3 + b3 + c3 ≥ a2 b + b2 c + c2 a
...
4 A wonderful inequality

19

Exercise 1
...
Any three positive real numbers a, b and c, with abc = 1, satisfy
a3 + b3 + c3 + (ab)3 + (bc)3 + (ca)3 ≥ 2(a2 b + b2 c + c2 a)
...
62
...

2
b
c
a
a b
c
Exercise 1
...
Any three positive real numbers a, b and c satisfy
1
1
1
a+b+c

...
64
...

b+c−a c+a−b a+b−c
Exercise 1
...
If a1 , a2 ,
...

+
+ ···+
≥
s − a1
s − a2
s − an
n−1
Exercise 1
...
If a1 , a2 ,
...

+
+ ···+
≥
s − a1
s − a2
s − an
n−1
Exercise 1
...
If a1 , a2 ,
...

+
+ ··· +
≥
2 − a1
2 − a2
2 − an
2n − 1
Exercise 1
...
(Quadratic mean-arithmetic mean inequality) Let x1 ,
...

n
n
Exercise 1
...
For positive real numbers a, b, c such that a + b + c = 1, prove that
ab + bc + ca ≤

1

...
70
...
, xn ∈ R+ ,
prove that
1
x1

+

1
x2

n
+ ···+

1
xn

≤

√
x1 + x2 + · · · + xn
n
x1 x2 · · · xn ≤

...

20

Numerical Inequalities

Exercise 1
...
Let a1 , a2 ,
...
Prove
that
1
1
1
an−1 + an−1 + · · · + an−1 ≥
+
+ ···+

...
72
...
, an be positive numbers such that
a1 + a2 + · · · + an = 1
...

+ ··· + √
≥ √
1 − a1
1 − an
n−1
Exercise 1
...
Let a, b and c be positive numbers such that a + b + c = 1
...

Exercise 1
...
Let a, b, c, d ∈ R+ with ab + bc + cd + da = 1, prove that
b3
c3
d3
1
a3
+
+
+
≥
...
75
...

b
c a
Exercise 1
...
Let x1 , x2 ,
...
Set s = n xk
...

s − 2xk
n−2
k=1

1
...

(1
...

1
...

L(s) = f (x) +
y−x
Then, evaluating at the point s = ty + (1 − t)x, we get
f (y) − f (x)
(t(y − x)) = f (x) + t(f (y) − f (x))
y−x
= tf (y) + (1 − t)f (x)
...
4) is equivalent to
f (ty + (1 − t)x) ≤ L(ty + (1 − t)x)
...
5
...
(1) If f is convex in the interval [a, b], then it is convex in
any subinterval [x, y] ⊂ [a, b]
...

2

(1
...
, tn ∈ [0, 1],
n
with i=1 ti = 1, and for x1 ,
...

(4) In particular, for x1 ,
...

n

22

Numerical Inequalities

Proof
...

(2) It is suﬃcient to choose t = 1 in (1
...

2
(3) We have
f (t1 x1 + · · · + tn xn ) = f ((1 − tn ) (
≤ (1 − tn ) f
≤ (1 − tn )

t1
tn−1
x1 + · · · +
xn−1 ) + tn xn )
1 − tn
1 − tn

t1
tn−1
x1 + · · · +
xn−1 + tn f (xn ), by convexity
1 − tn
1 − tn
tn−1
t1
f (x1 ) + · · · +
f (xn−1 ) + tn f (xn ), by induction
1 − tn
1 − tn

= t1 f (x1 ) + · · · + tn f (xn )
...

Observations 1
...
2
...

2
2
(ii) We can observe that (3) is true for t1 ,
...

We will prove (i) using induction
...
, xn ∈ [a, b]
...

Now, we will show that Pn ⇒ Pn−1
...
, xn ∈ [a, b] and let y = x1 +···+xn−1
...

n
n
n

But the left side is f (y), therefore n · f (y) ≤ f (x1 ) + · · · + f (xn−1 ) + f (y), and
f (y) ≤

1
(f (x1 ) + · · · + f (xn−1 ))
...

Let D = f
v=

xn+1 +···+x2n

...
5 Convex functions

23

where we have used twice the statement that Pn is true
...
, xn ∈ [a, b] and n ∈ N
...
, tn = rn be rational numbers in [0, 1] with i=1 ti = 1
...

Observation 1
...
3
...

We have seen that if f satisﬁes (2), then
f (qx + (1 − q)y) ≤ qf (x) + (1 − q)f (y)
for any x, y ∈ [a, b] and q ∈ [0, 1] rational number
...

Now, by using the continuity of f and taking the limit, we get
f (tx + (1 − t)y) ≤ tf (x) + (1 − t)f (y)
...

2A

function f : [a, b] → R is continuous at a point c ∈ [a, b] if lim f (x) = f (c), and f is
x→c

continuous on [a, b] if it is continous in every point of the interval
...

24

Numerical Inequalities

Observation 1
...
4
...

Now, we will consider some criteria to decide whether a function is convex
...
5
...
A function f : [a, b] → R is convex if and only if the set {(x, y)|
a ≤ x ≤ b, f (x) ≤ y} is convex
...
Suppose that f is convex and let A = (x1 , y1 ) and B = (x2 , y2 ) be two
points in the set U = {(x, y) | a ≤ x ≤ b, f (x) ≤ y}
...
The ﬁrst condition
follows immediately since x1 and x2 belong to [a, b]
...

Moreover, since f (x2 ) ≤ y2 and f (x1 ) ≤ y1 , we can deduce that
f (tx2 + (1 − t)x1 ) ≤ ty2 + (1 − t)y1
...

Let x1 , x2 ∈ [a, b] and let us consider A = (x1 , f (x1 )) and B = (x2 , f (x2 ))
...
Thus,
(tx2 + (1 − t)x1 , tf (x2 ) + (1 − t)f (x1 )) ∈ U ,
but this implies that f (tx2 + (1 − t)x1 ) ≤ tf (x2 ) + (1 − t)f (x1 )
...

Criterion 1
...
6
...

x−x
Proof
...
To prove that P (x) is non-decreasing, we take
x < y and then we show that P (x) ≤ P (y)
...
Let us consider the ﬁrst of these
3 A subset C of the plane is convex if for any pair of points A, B in C, the segment determined
by these points belongs entirely to C
...

1
...
First note that
f (y) − f (x0 )
f (x) − f (x0 )
≤
x − x0
y − x0
⇔ (f (x) − f (x0 ))(y − x0 ) ≤ (f (y) − f (x0 ))(x − x0 )

P (x) ≤ P (y) ⇔

⇔ f (x)(y − x0 ) ≤ f (y)(x − x0 ) + f (x0 )(y − x)
x − x0
y−x
⇔ f (x) ≤ f (y)
+ f (x0 )
y − x0
y − x0
x − x0
x − x0
y−x
y−x
⇔ f
y+
x0 ≤ f (y)
+ f (x0 )

...

Criterion 1
...
7
...
In particular, if f is twice diﬀerentiable
and f (x) ≥ 0, then the function is convex
...
It is clear that f (x) ≥ 0, for x ∈ [a, b], implies that f (x) is non-decreasing
...

Let x = tb + (1 − t)a be a point on [a, b]
...

Then, since f (x) is non-decreasing, we can deduce that
(1 − t) (f (x) − f (a)) = t(1 − t)(b − a)f (c) ≤ t(1 − t)(b − a)f (d) = t(f (b) − f (x))
...

Let us present one geometric interpretation of convexity (and concavity)
...
If the vertices of
the triangle XY Z have coordinates X = (x, f (x)), Y = (y, f (y)), Z = (z, f (z)),
then the area of the triangle is given by
⎞
⎛
1 x f (x)
1
Δ = det A, where A = ⎝ 1 y f (y) ⎠
...

f (x)−f (c)
exists and f is diﬀerentiable in A ⊂
x−c
x→c
5 Mean value theorem
...
See [21, page 169]
...
For a
convex function, we have that Δ > 0 and for a concave function, Δ < 0, as shown
in the following graphs
...

z−x
z−x

If we take t = y−x , we have 0 < t < 1, 1 − t =
z−x
f (tz + (1 − t)x) < tf (z) + (1 − t)f (x)
...

Example 1
...
8
...

This follows from the fact that f (x) = n(n − 1)xn−2 ≥ 0 in each case
...

2

2

2

2

2

a +b
≤ a +b , we can deduce that a+b ≤
(i) Since a+b
2
2
2
2 , which is the
inequality between the arithmetic mean and the quadratic mean
...

(iii) If a and b are positive numbers, 1 +

a n
b

+ 1+

b n
a

≥ 2n+1
...
5 Convex functions

27

from
a+b
a

+
2

a+b
b

≤

1
a
f 1+
+f
2
b

=

2n = f (2) ≤ f

1
2

1+

a
b

n

1+

+ 1+

b
a

b
a
n

...
5
...
The exponential function f (x) = ex is convex in R, since f (x) =
ex > 0, for every x ∈ R
...

(i) (Weighted AM-GM inequality) If x1 ,
...
, tn are positive numbers
n
and i=1 ti = 1, then
xt1 · · · xtn ≤ t1 x1 + · · · + tn xn
...

1
In particular, if we take ti = n , for 1 ≤ i ≤ n, we can produce another proof
of the inequality between the arithmetic mean and the geometric mean for n
numbers
...
If a, b > 0 satisfy the
1
1
condition a + 1 = 1, then xy ≤ a xa + 1 y b
...

a
b

(iii) (H¨lder’s inequality) Let x1 , x2 ,
...
, yn be positive numbers
o
1
and a, b > 0 such that a + 1 = 1, then
b
n

1/a

n

xi yi ≤
i=1

b
yi

i=1

n

xi yi ≤
i=1

1
a

n
n
a
i=1 xi =
i=1
1 a
1 b
a xi + b yi , then
n

xa +
i
i=1

...

n
b
yi =
i=1

1 1
+ = 1
...
Since
1/b
n

n
i=1

a

(xi ) =

xa
i

A

i=1

n
i=1

xa = A and
i

b
yi = B
...

i=1

xi yi ≤ A1/a B 1/b
...

Let us introduce a consequence of H¨lder’s inequality, which is a generalizao
tion of the triangle inequality
...
5
...
Let a1 , a2 ,
...
, bn be
positive numbers and p > 1, then
1
p

n

≤

(ak + bk )p

1
p

n

k=1

(ak )p

1
p

n

(bk )p

+

k=1

...

(1
...
6), with q such that 1 + 1 = 1, to get
p
q
n

1
p

n
p−1

ak (ak + bk )

≤

q(p−1)

(ak )

k=1
n

n

(ak + bk )

k=1

bk (ak + bk )p−1 ≤
k=1

1
q

n

p
1
p

(bk )p
k=1

,

k=1
1
q

n

(ak + bk )q(p−1)

...
6), and noting that q(p − 1) = p, yields the
required inequality
...
For 0 1, then ri = exi for some xi > 0
...

√
n
r1 · · · rn + 1
1 + r1
1 + rn

Example 1
...
12
...
,
n
xn such that i=1 xi = 1, we have
n
i=1

n √
xi
x
i=1
√ i

...

≥√
1 − xi
n−1
i=1
√
n √
It is left to prove that i=1 xi ≤ n, but this follows from the Cauchy-Schwarz
√
√
n
n
inequality, n
xi ≤
n
...
5
...
(Hungary–Israel, 1999) Let k and l be two given positive integers,
and let aij , 1 ≤ i ≤ k and 1 ≤ j ≤ l, be kl given positive numbers
...

ij

30

Numerical Inequalities
k

Deﬁne bj = i=1 ap for j = 1, 2,
...
Then
l

q

Lq =

p
bj

j=1
l
j=1
k
i=1

l

⎢⎝
⎣

i=1

bj

=

⎞ q−p ⎛
q
⎠

⎝

l

⎞ ⎤
⎥
aq ⎠ ⎦
ij

⎛

q−p
q

q
p

bj ⎠

⎝

j=1
l

⎞p ⎤
q
q
p p⎠ ⎥
(aij )
⎦

l
j=1

⎞

l

⎞

ap ⎠
...

ij

j=1

The inequality L ≤ R follows by dividing both sides of Lq ≤ Lq−p Rp by Lq−p and
taking the p-th root
...
77
...

2

(ii) For a, b, c ∈ R+ , with a + b + c = 1, prove that
a+

1
a

2

+ b+

1
b

2

+ c+

1
c

2

≥

100

...
78
...

b+c+1 c+a+1 a+b+1

1
...
79
...

+
≤ √
2
2
1 + xy
1+x
1+y
Exercise 1
...
Prove that the function f (x) = sin x is concave in the interval [0, π]
...

Exercise 1
...
If A, B, C, D are angles belonging to the interval [0, π], then
(i) sin A sin B ≤ sin2

A+B
2

and the equality holds if and only if A = B,

(ii) sin A sin B sin C sin D ≤ sin4
(iii) sin A sin B sin C ≤ sin3

A+B+C+D
4

A+B+C
3

,

,

Moreover, if A, B, C are the internal angles of a triangle, then
√
(iv) sin A sin B sin C ≤ 3 3,
8
(v) sin A sin B sin C ≤ 1 ,
2
2
2
8
(vi) sin A + sin B + sin C = 4 cos A cos B cos C
...
82
...

(ii) Use this inequality to provide another proof of the AM-GM inequality
...
83
...

For the case n = 1, the inequality can take one of the following forms:
(a) x3 + y 3 + z 3 + 3xyz ≥ xy(x + y) + yz(y + z) + zx(z + x)
...

(c) If x + y + z = 1, 9xyz + 1 ≥ 4(xy + yz + zx)
...
84
...

Exercise 1
...
If a, b, c are positive real numbers, prove that
b
c
9
a

...
86
...

ab + bc + ca
a+b+c

Moreover, if abc = 1, prove that
1+

6
3
≥

...
87
...
, xn be positive real numbers
and let t1 , t2 ,
...
Let r and s be two
nonzero real numbers such that r > s
...

Exercise 1
...
(Two extensions of H¨lder’s inequality) Let x1 , x2 ,
...
, yn , z1 , z2 ,
...

(i) If a, b, c are positive real numbers such that
1
c

n

(xi yi )

xi

i=1

n

xi yi zi ≤
i=1

yi
1
a

+
1
b

n

xi a
i=1

1
b

n
b

...

i=1

Exercise 1
...
(Popoviciu’s inequality) If I is an interval and f : I → R is a convex
function, then for a, b, c ∈ I the following inequality holds:
2
f
3

a+b
2

+f

b+c
2

+ f

c+a
2
≤

f (a) + f (b) + f (c)
+f
3

a+b+c
3

Exercise 1
...
Let a, b, c be non-negative real numbers
...

Exercise 1
...
Let a, b, c be positive real numbers
...

...
6 A helpful inequality

33

1
...

Let P denote the cubic polynomial
P (x) = x3 − (a + b + c)x2 + (ab + bc + ca)x − abc,
which has a, b and c as its roots
...

Adding up these three equations yields
a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca)
...
7)

It immediately follows that if a + b + c = 0, then a3 + b3 + c3 = 3abc
...

2

(1
...
7),
a3 + b3 + c3 − 3abc =

1
(a + b + c)[(a − b)2 + (b − c)2 + (c − a)2 ]
...
9)

This presentation of the identity leads to a short proof of the AM-GM inequality
for three variables
...
9) it is clear that if a, b, c are positive numbers, then
√
√
a3 + b3 + c3 ≥ 3abc
...

Note that identity (1
...
27
...
92
...

34

Numerical Inequalities

Exercise 1
...
For positive real numbers a, b, c, prove that
1 1 1
a 2 + b 2 + c2
≥ + +
...
94
...

Exercise 1
...
Let a, b, c be the side lengths of a triangle
...

2

Exercise 1
...
(Romania, 2007) For non-negative real numbers x, y, z, prove that
x3 + y 3 + z 3
3
≥ xyz + |(x − y)(y − z)(z − x)|
...
97
...

A very simple inequality which may be helpful for proving a large number of
algebraic inequalities is the following
...
6
...
If a, b, x, y are real numbers and x, y > 0,
then the following inequality holds:
a2
b2
(a + b)2
+
≥

...
10)

Proof
...
Clearing out denominators, we can express the
inequality as
a2 y(x + y) + b2 x(x + y) ≥ (a + b)2 xy,
which simpliﬁes to become the obvious (ay − bx)2 ≥ 0
...

Another form to prove the inequality is using the Cauchy-Schwarz inequality
in the following way:
(a + b)2 =

a √
b √
√
x+ √ y
x
y

2

≤

b2
a2
+
x
y

(x + y)
...
6 A helpful inequality

35

and a simple inductive argument shows that
a2
a2
(a1 + a2 + · · · + an )2
a2
1
+ 2 + ··· + n ≥
x1
x2
xn
x1 + x2 + · · · + xn

(1
...
, an and x1 , x2 ,
...

x1
x2
xn
Inequality (1
...

As a ﬁrst application of this inequality, we will present another proof of the
Cauchy-Schwarz inequality
...

2 + b2 + · · · + b2
b1
b2 + b2 + · · · + b2
n
n
2
1
2

Thus, we conclude that
(a2 + a2 + · · · + a2 )(b2 + b2 + · · · + b2 ) ≥ (a1 b1 + a2 b2 + · · · + an bn )2
1
2
n
1
2
n
and the equality holds if and only if
a1
a2
an
=
= ··· =

...

Example 1
...
2
...
, an , b1 ,
...
Prove that
(i)

an
(a1 + · · · + an )2
a1
+ ···+
≥
,
b1
bn
a 1 b 1 + · · · + an b n

(ii)

an
1
a1
2 + · · · + b2 ≥ a + · · · + a
b1
1
n
n

a1
an
+ ··· +
b1
bn

2

...
11), as we can see
as follows
...

36

Numerical Inequalities

Example 1
...
3
...
, an , b1 ,
...
Prove that
a2
a2
1
n
1
+ ···+
≥ (a1 + · · · + an )
...
11) implies that
a2
a2
(a1 + a2 + · · · + an )2
1
n
+ ···+
≥
a1 + b 1
an + b n
a 1 + a2 + · · · + an + b 1 + b 2 + · · · + b n
=

(a1 + a2 + · · · + an )2
2(a1 + a2 + · · · + an )

=

1
(a1 + a2 + · · · + an )
...

Example 1
...
4 (Quadratic mean-arithmetic mean inequality)
...
, xn , we have
x2 + x2 + · · · + x2
x1 + x2 + · · · + xn
1
2
n
≥

...
11) leads us to
x2 + x2 + · · · + x2
(x1 + x2 + · · · + xn )2
1
2
n
≥
,
n
n2
which implies the above inequality
...
Our next application
shows this trick and oﬀers a shorter proof for Example 1
...
9
...
6
...
(IMO, 1995) Let a, b, c be positive real numbers such that abc = 1
...

a3 (b + c) b3 (a + c) c3 (a + b)
2
Observe that
1
1
1
1
1
1
a2
b2
c2
+ 3
+ 3
=
+
+
a3 (b + c) b (c + a) c (a + b)
a(b + c) b(c + a) c(a + b)

≥

( 1 + 1 + 1 )2
ab + bc + ca
a
b
c
=
2(ab + bc + ca)
2(abc)

≥

3 3 (abc)2
3
= ,
2
2

1
...
11) and the second is a consequence of
the AM -GM inequality
...
11), we provide a simple
proof of Nesbitt’s inequality
...
6
...
For a, b, c ∈ R+ , we have
b
c
3
a
+
+
≥
...
11) to produce

a b
a, b,

a2
b2
c2
(a + b + c)2
+
+
≥

...
8) we know that a2 + b2 + c2 − ab − bc − ca ≥ 0, that is,
(a + b + c)2 ≥ 3(ab + bc + ca)
...

b+c c+a a+b
2(ab + bc + ca)
2
Example 1
...
7
...

b + 2c c + 2a a + 2b
Observe that
a
b
c
a2
b2
c2
+
+
=
+
+

...
11) yields
b2
c2
(a + b + c)2
a2
+
+
≥
≥ 1,
ab + 2ca bc + 2ab ca + 2bc
3(ab + bc + ca)
where the last inequality follows in the same way as in the previous example
...
98
...

a b
c
d
a+b+c+d
Exercise 1
...
Let a and b be positive real numbers
...

38

Numerical Inequalities

Exercise 1
...
Let x, y, z be positive real numbers
...

x+y y+z
z+x
x+y+z
Exercise 1
...
Let a, b, x, y, z be positive real numbers
...

ay + bz az + bx ax + by
a+b
Exercise 1
...
Let a, b, c be positive real numbers
...

a+b
b+c
c+a
Exercise 1
...

(i) Let x, y, z be positive real numbers
...

x + 2y + 3z y + 2z + 3x z + 2x + 3y
2

(ii) (Moldova, 2007) Let w, x, y, z be positive real numbers
...

x + 2y + 3z y + 2z + 3w z + 2w + 3x w + 2x + 3y
3
Exercise 1
...
(Croatia, 2004) Let x, y, z be positive real numbers
...

(x + y)(x + z) (y + z)(y + x) (z + x)(z + y)
4
Exercise 1
...
For a, b, c, d positive real numbers, prove that
a
b
c
d
+
+
+
≥ 2
...
106
...
Prove that
b
c
d
e
5
a
+
+
+
+
≥
...
107
...

x
y
z
3(x + y + z)
(ii) (Belarus, 2000) Prove that, for all positive real numbers a, b, c, x, y, z, the
following inequality holds:
a3
b3
c3
(a + b + c)3
+
+
≥

...
7 The substitution strategy

39

Exercise 1
...
(Greece, 2008) For x1 , x2 ,
...
, xn } and t = min {x1 , x2 ,
...
Under which condition the equality holds?

1
...
Making an adequate
substitution we can, for instance, change the diﬃcult terms of the inequality a
little, we can simplify expressions or we can reduce terms
...
As always, the best way to do
that is through some examples
...
In the next example we
apply this technique to eliminate the denominators in order to make the problem
easier to solve
...
7
...
If a, b, c are positive real numbers less than 1, with a + b + c = 2,
then
a
b
c
≥ 8
...
Hence
the inequality is equivalent to
y+z
x

z+x
y

x+y
z

≥ 8,

and in turn, this is equivalent to
(x + y)(y + z)(z + x) ≥ 8xyz
...
It is enough to apply three times the
√
AM-GM inequality under the form (x + y) ≥ 2 xy (see Exercise 1
...

It may be possible that the extra condition is used only as part of the solution,
as in the following two examples
...
7
...
(Mexico, 2007) If a, b, c are positive real numbers that satisfy
a + b + c = 1, prove that
√
√
√
a + bc + b + ca + c + ab ≤ 2
...

2

Similarly,
√
2b + c + a
b + ca ≤
2

and

√
2c + a + b

...

≤
2
2
2
2
The equality holds when a + b = a + c, b + c = b + a and c + a = c + b, that is,
when a = b = c = 1
...
7
...
If a, b, c are positive real numbers with ab + bc + ca = 1, prove
that
a
b
c
3
√
+√
+√
≤
...
Similarly, b2 + 1 =
(b + c)(b + a) and c2 + 1 = (c + a)(c + b)
...

2

Using the AM-GM inequality, applied to every element of the sum on the left-hand
side, we obtain
a

+

b

+

c

(a + b)(a + c)
(b + c)(b + a)
(c + a)(c + b)
a
a
1
b
b
1
c
c
1
+
+
+
+
+
≤
2 a+b a+c
2 b+c b+a
2 c+a c+b

=

3

...
In the
following example the substitution allows us to make at least one of the terms in
the inequality look simpler
...
7
...
(India, 2002) If a, b, c are positive real numbers, prove that
a b
c
c+a a+b
b+c
+ + ≥
+
+

...
7 The substitution strategy

41

c
Making the substitution x = a , y = b , z = a the left-hand side of the
b
c
inequality is now more simple, x + y + z
...
The ﬁrst element of the sum is modiﬁed as follows:

1+
c+a
=
c+b
1+

a
c
b
c

=

ab
b c
+b
c

1+
1

1−x
1 + xy
=x+

...

a+c
1+z
b+a
1+x
Now, the inequality is equivalent to
x−1 y−1 z−1
+
+
≥0
1+y
1+z
1+x

with the extra condition xyz = 1
...

But, from the AM-GM inequality, we have x2 z + y 2 x+ z 2 y ≥ 3 3 x3 y 3 z 3 = 3
...
11)
...
This example also helps us
to point out that we may need to make more than one substitution in the same
problem
...
7
...
Let a, b, c be positive real numbers, prove that
(a + b)(a + c) ≥ 2

abc(a + b + c)
...

1
1
Now, dividing both sides by xy and making the substitution r = 1 + x , s = 1 + y ,
the inequality we need to prove becomes
√
rs ≥ 2 rs − 1
...

It is a common situation for inequality problems to have several solutions
and also to accept several substitutions that help to solve the problem
...

42

Numerical Inequalities

Example 1
...
6
...

2
2
2
2
1+a
1+b
1+c
1
Under the substitution x = a , y = 1 , z = 1 , condition a + b + c = abc
b
c
becomes xy + yz + zx = 1 and the inequality becomes equivalent to

√

x
x2

+1

+

y
y2

3
z
≤
...

Another solution is to make the substitution a = tan A, b = tan B, c = tan C
...
Now, since 1 + tan2 A = (cos A)−2 , the inequality is equivalent to
cos A + cos B + cos C ≤ 3 , which is a valid result as will be shown in Example
2
2
...
2
...

We note that not all substitutions are algebraic, since there are trigonometric
substitutions that can be useful, as is shown in the last example and as we will
see next
...
2 and 2
...

Example 1
...
7
...

Making the substitution a = cos2 A, b =√ 2 B, c = cos2 C, with A, B, C in
cos
√
√
the interval (0, π ), we obtain that 1 − a = 1 − cos2 A = sin A, 1 − b = sin B
2
√
and 1 − c = sin C
...

But observe that
cos A cos B cos C + sin A sin B sin C < cos A cos B + sin A sin B
= cos(A − B) ≤ 1
...
109
...
Prove that
x3

x3
y3
z3
+ 3
+ 3
≥ 1
...
110
...
Prove that
1
1
1
3
+
+
≥
...
8 Muirhead’s theorem

43

Exercise 1
...
(Russia, 2004) If n > 3 and x1 , x2 ,
...

1 + x1 + x1 x2
1 + x2 + x2 x3
1 + xn + xn x1
Exercise 1
...
(Poland, 2006) Let a, b, c be positive real numbers such that
ab + bc + ca = abc
...

ab(a3 + b3 ) bc(b3 + c3 ) ca(c3 + a3 )
Exercise 1
...
(Ireland, 2007) Let a, b, c be positive real numbers, prove that
1
3

bc ca ca
+
+
a
b
b

a 2 + b 2 + c2
a+b+c
≥

...
114
...

Prove that
a−2 b−2 c−2
+
+
≤ 0
...
8 Muirhead’s theorem
In 1903, R
...
Muirhead published a paper containing the study of some algebraic
methods applicable to identities and inequalities of symmetric algebraic functions
of n variables
...
, an ) satisfying the condition
a1 ≥ a2 ≥ · · · ≥ a n
...
We will denote by
!

F (x1 ,
...
, xn )
...
, xn ) = xa1 xa2 · · · xan
n
1 2
We write [a] = [a1 , a2 ,
...

xa1 xa2 · · · xan
...

3!

44

Numerical Inequalities

It is clear that [a] is invariant under any permutation of the (a1 , a2 ,
...
We will
say that a mean value of the type [a] is a symmetrical mean
...
, 0] = (n−1)! (x1 + x2 + · · · + xn ) = n
i=1 xi is the arithmetic mean
n!
1
1
1
√
1 1
1
n!
n
n
n
n
and [ n , n ,
...

When a1 + a2 + · · · + an = 1, [a] is a common generalization of both the arithmetic
mean and the geometric mean
...
, xn
...

From now on we denote (a) = (a1 , a2 ,
...

Deﬁnition 1
...
1
...

i=1

It is clear that (a) ≺ (a) and that (b) ≺ (a) and (c) ≺ (b) implies (c) ≺ (a)
...
8
...
[b] ≤ [a] for any n-tuple of non-negative
numbers (x1 , x2 ,
...
Equality takes place only when
(b) and (a) are identical or when all the xi s are equal
...
First, it is clear that [2, 0, 0] cannot be compared with [1, 1, 1] because
the ﬁrst condition in Deﬁnition 1
...
1 is not satisﬁed, but we can see that [2, 0, 0] ≥
[1, 1, 0], which is equivalent to
x2 + y 2 + z 2 ≥ xy + yz + zx
...
x2 + y 2 ≥ 2xy ⇔ [2, 0] ≥ [1, 1],
2
...
x5 + y 5 ≥ x3 y 2 + x2 y 3 ⇔ [5, 0] ≥ [3, 2],
4
...
8 Muirhead’s theorem

45

and all these inequalities are satisﬁed if we take for granted Muirhead’s theorem
...
Suppose that [b] ≤ [a] for any n positive numbers
x1 , x2 ,
...
Taking xi = x, for all i, we obtain
x

bi

= [b] ≤ [a] = x

ai

...

Next, take x1 = x2 = · · · = xν = x, xν+1 = · · · = xn = 1 and x very large
...
Thus, it is clear that the ﬁrst sum can not be greater than the second
and this proves (2) in Deﬁnition 1
...
1
...

We deﬁne a special type of linear transformation T of the a’s, as follows
...

If now 0 ≤ σ < τ ≤ ρ, then a T -transformation is deﬁned by
τ +σ
τ −σ
ak +
al ,
2τ
2τ
τ −σ
τ +σ
ak +
al ,
T (al ) = bl = ρ − σ =
2τ
2τ
T (aν ) = aν (ν = k, ν = l)
...
The deﬁnition does
not necessarily imply that either the (a) or the (b) are in decreasing order
...

Lemma 1
...
3
...

Proof
...
Thus
[a] − [b] = [ρ + τ, ρ − τ, a3 ,
...
]
1
xa3 · · · xan (xρ+τ xρ−τ + xρ−τ xρ+τ )
=
n
1
2
1
2
! 3
2n!
1
−
xa3 · · · xan (xρ+σ xρ−σ + xρ−σ xρ+σ )
n
1
2
1
2
! 3
2n!
1
=
(x1 x2 )ρ−τ xa3 · · · xan (xτ +σ − xτ +σ )(xτ −σ − xτ −σ ) ≥ 0
n
3
1
2
1
2
!
2n!
with equality being the case only when all the xi ’s are equal
...
8
...
If (b) ≺ (a), but (b) is not identical to (a), then (b) can be derived
from (a) using the successive application of a ﬁnite number of T -transformations
...
We call the number of diﬀerences aν −bν which are not zero, the discrepancy
between (a) and (b)
...
We will prove
the lemma by induction, assuming it to be true when the discrepancy is less than
r and proving that it is also true when the discrepancy is r
...
Since
n
n
(aν − bν ) = 0, and not all of these diﬀerences are zero,
i=1 ai =
i=1 bi , and
there must be positive and negative diﬀerences, and the ﬁrst which is not zero
must be positive because of the second condition of (b) ≺ (a)
...
, bl−1 = al−1 , bl > al ;

(1
...

We take ak = ρ + τ , al = ρ − τ , and deﬁne σ by
σ = max(|bk − ρ| , |bl − ρ|)
...
Also, one (possible both) of bl − ρ = −σ or
bk − ρ = σ is true, since bk ≥ bl , and σ < τ , since bk < ak and bl > al
...

We now write ak = ρ + σ, al = ρ − σ, aν = aν (ν = k, ν = l)
...
Since the pairs ak , bk and al , bl each
contributes one unit to the discrepancy r between (b) and (a), the discrepancy
between (b) and (a ) is smaller, being equal to r − 1 or r − 2
...

Finally, let us prove that (b) ≺ (a )
...

For the ﬁrst one, we have
n

ak + al = 2ρ = ak + al ,

n

bi =
i=1

n

ai =
i=1

ai
...

Now, this is true if ν < k or ν ≥ l, as can be established by using the deﬁnition of
(a ) and also the second condition of (b) ≺ (a)
...

1
...
12),
ak−1 = ak−1 ≥ ak = ρ + τ > ρ + σ = ak ≥ bk ≥ bk+1 = ak+1 = ak+1 ,
al−1 = al−1 = bl−1 ≥ bl ≥ al = ρ − σ > ρ − τ = al ≥ al+1 = al+1
...

We have thus proved that (b) ≺ (a ), a set arising from (a) using a transformation T and having a discrepancy from (b) of less than r
...

The proof of Muirhead’s theorem demonstrates to us how the diﬀerence between two comparable means can be decomposed as a sum of obviously positive
terms by repeated application of the T -transformation
...

Example 1
...
5 (The AM-GM inequality)
...
,
yn ,
y1 + y2 + · · · + yn
√
≥ n y1 y2 · · · yn
...

Now, we observe that
1
n

n

xn = [n, 0, 0,
...
, 1]
...
, 0] ≥ [1, 1,
...

Next, we provide another proof for the AM-GM inequality, something we
shall do by following the ideas inherent in the proof of Muirhead’s theorem in

48

Numerical Inequalities

order to illustrate how it works
...
, 0] − [1, 1,
...
, 0] − [n − 1, 1, 0,
...
, 0] − [n − 2, 1, 1, 0,
...
, 0] − [n − 3, 1, 1, 1, 0,
...
, 1] − [1, 1,
...

1
2

Since (xν − xν )(xr − xs ) > 0, unless xr = xs , the inequality follows
...
8
...
If a, b are positive real numbers, then
a2
+
b

Setting x =

√
√
b2
≥ a + b
...

Using Muirhead’s theorem, we get
[3, 0] =

1
1 3
(x + y 3 ) ≥ xy(x + y) = [2, 1],
2
2

and thus the result follows
...
8
...
If a, b, c are non-negative real numbers, prove that
a3 + b3 + c3 + abc ≥

1
(a + b + c)3
...

Then we need to prove that
3[3, 0, 0] + 6[1, 1, 1] ≥

1
(3[3, 0, 0] + 18[2, 1, 0] + 36[1, 1, 1]),
7

that is,
36
18
[3, 0, 0] + 6 −
7
7

[1, 1, 1] ≥

18
[2, 1, 0]
7

1
...

This follows using the inequalities [3, 0, 0] ≥ [2, 1, 0] and [1, 1, 1] ≥ 0
...
8
...
If a, b, c are non-negative real numbers, prove that
a+b+c≤

a2 + b2
b 2 + c2
c2 + a2
a3
b3
c3
+
+
≤
+
+
...
Using Muirhead’s theorem we arrive at the result
...
115
...

Exercise 1
...
(IMO, 1961) Let a, b, c be the lengths of the sides of a triangle,
and let (ABC) denote its area
...

Exercise 1
...
Let a, b, c be positive real numbers
...

(a + b)(a + c) (b + c)(b + a) (c + a)(c + b)
4(a + b + c)
Exercise 1
...
(IMO, 1964) Let a, b, c be positive real numbers
...

Exercise 1
...
(Short list Iberoamerican, 2003) Let a, b, c be positive real numbers
...

2
2
− bc + c
c − ca + a
a − ab + b2

Exercise 1
...
(Short list IMO, 1998) Let a, b, c be positive real numbers such
that abc = 1
...

(1 + b)(1 + c) (1 + c)(1 + a) (1 + a)(1 + b)
4

Chapter 2

Geometric Inequalities
2
...
One of them is the triangle inequality and we will refer to it as D1; the
second one is not really an inequality, but it represents an important observation
concerning the geometry of triangles which points out that if we know the greatest
angle of a triangle, then we know which is the longest side of the triangle; this
observation will be denoted as D2
...
If A, B and C are points on the plane, then
AB + BC ≥ AC
...

D2
...

Hence, if in the triangle ABC we have ∠A > ∠B, then BC > CA
...
1
...

(ii) To be able to construct a triangle with side lengths a ≤ b ≤ c, it is suﬃcient
that c < a + b
...

Exercise 2
...
(i) If it is possible to construct a triangle with side-lengths a < √<
b
√
√
c, then it is possible to construct a triangle with side-lengths a < b < c
...

52

Geometric Inequalities

(iii) If it is possible to construct a triangle with side-lengths a < b < c, then it is
1
1
1
possible to construct a triangle with side-lengths a+b , b+c and c+a
...
3
...
Prove that there are three
of them that form an acute triangle
...

Example 2
...
1
...

Since c2 = a2 + b2 − ab = a2 + b2 − 2ab cos 60◦ , we can think that a, b, c are
the lengths of the sides of a triangle such that the measure of the angle opposed
to the side of length c is 60◦
...
In any case it follows that (a − b)(b − c) ≤ 0
...
1
...
We can also solve the example above without the identiﬁcation
of a, b and c with the lengths of the sides of a triangle
...

Similarly, a ≥ b implies c − b ≥ 0, and hence
(a − b)(b − c) ≤ 0
...

Example 2
...
3
...

The radicals suggest using the cosine law with angles of 120◦ and of 60◦ as
follows: a2 + ac + c2 = a2 + c2 − 2ac cos 120◦ , a2 − ab + b2 = a2 + b2 − 2ab cos 60◦
and b2 − bc + c2 = b2 + c2 − 2bc cos 60◦
...
1 Two basic inequalities

53

Then, if we consider a quadrilateral ABCD, with ∠ADB = ∠BDC = 60◦ and
∠ADC = 120◦, such that AD = a, BD = b and CD = c, we can deduce that AB =
√
√
√
a2 − ab + b2 , BC = b2 − bc + c2 and CA = a2 + ac + c2
...

Exercise 2
...
Let ABC be a triangle with ∠A > ∠B, prove that BC > 1 AB
...
5
...

2
Exercise 2
...
If a1 , a2 , a3 , a4 and a5 are the lengths of the sides of a convex
pentagon and if d1 , d2 , d3 , d4 and d5 are the lengths of its diagonals, prove that
1
a 1 + a2 + a3 + a4 + a5
<
< 1
...
7
...

2
Exercise 2
...
If the length ma of the median AA of a triangle ABC satisﬁes
ma > 1 a, prove that ∠BAC < 90◦
...
9
...

Exercise 2
...
If ma , mb and mc are the lengths of the medians of a triangle with
side-lengths a, b and c, respectively, prove that it is possible to construct a triangle
with side-lengths ma , mb and mc , and that
3
(a + b + c) < ma + mb + mc < a + b + c
...
11
...
The equality holds if and only if ABCD is a
cyclic quadrilateral
...
12
...
Prove that AC > BD if and
only if (AD − BC)(AB − DC) > 0
...
13
...
Prove that P A,
P B and P C are the lengths of the sides of a triangle
...
14
...

54

Geometric Inequalities

Exercise 2
...
If a, b and c are the lengths of the sides of a triangle, ma , mb and
mc represent the lengths of the medians and R is the circumradius, prove that
(i)

a2 + b 2
b 2 + c2
c2 + a2
+
+
≤ 12R,
mc
ma
mb

(ii) ma (bc − a2 ) + mb (ca − b2 ) + mc (ab − c2 ) ≥ 0
...
16
...
Suppose
that c > b, prove that
1
3
(c − b) < mb − mc < (c − b),
2
2
where mb and mc are the lengths of the medians
...
17
...
Let D be the
intersection of the internal angle bisector of ∠A with the side BC and let Ia be
the center of the excircle of the triangle ABC opposite to the vertex A
...

DIa

2
...
One sort of problems consists of those where you are
asked to prove some inequality that is satisﬁed by the lengths of the sides of a
triangle without any other geometric elements being involved, as in the following
example
...
2
...
The lengths a, b and c of the sides of a triangle satisfy
a (b + c − a) < 2bc
...
We will prove the inequality in the following cases
...
a ≤ b
...

a

Case 2
...

In this case b − a ≤ 0, and since a 
2 a4 + b4 + c4 , then a, b and c are the lengths of the sides of a triangle
...

For part (i), it is suﬃcient to observe that
a 2 + b 2 + c2

2

− 2 a4 + b4 + c4 = (a + b + c)(a + b − c)(a − b + c)(−a + b + c) > 0,

and then note that none of these factors is negative
...
2
...

For part (ii), we can deduce that
3 a4 + b4 + c4 + d4 < a2 + b2 + c2 + d2
=

2

a 2 + b 2 + c2
a 2 + b 2 + c2
+
+ d2
2
2
a 2 + b 2 + c2
2

≤

2

+

a 2 + b 2 + c2
2

2

2

+ d4

√
3

2

...
Using the ﬁrst part we can deduce that a, b and c can be
b 4 + c4 < 2
4
used to construct a triangle
...

There is a technique that helps to transform one inequality between the
lengths of the sides of a triangle into an inequality between positive numbers (of
course related to the sides)
...

If the incircle (I, r) of the triangle ABC is tangent to the sides BC, CA
and AB at the points X, Y and Z, respectively, we have that x = AZ = Y A,
y = ZB = BX and z = XC = CY
...

2
y

Let us now see how to use the Ravi transformation in the following example
...
2
...
The lengths of the sides a, b and c of a triangle satisfy
(b + c − a)(c + a − b)(a + b − c) ≤ abc
...

Thus, the inequality is equivalent to
8xyz ≤ (x + y)(y + z)(z + x)
...
1)

The last inequality follows from Exercise 1
...

Example 2
...
4
...

If we set a = y + z, b = z + x, c = x + y, we can deduce that a + b − c = 2z,
b + c − a = 2x, c + a − b = 2y
...

Now applying the inequality between the arithmetic mean and the quadratic mean
(see Exercise 1
...

≤

Moreover, the equality holds if and only if x = y = z, that is, if and only if
a = b = c
...
Since a = x + y, b = y + z
and c = z + x, we ﬁrst obtain that s = a+b+c = x + y + z
...

The formula (ABC) = sr leads us to
r=

(ABC)
=
s

(x + y + z)xyz
=
x+y+z

xyz

...
2)

2
...

4 (x + y + z)xyz

Example 2
...
5
...
If
b
c
we construct a triangle A B C with side lengths a + 2 , b + 2 , c + a , prove that
2
9
(A B C ) ≥ 4 (ABC)
...
Using Heron’s formula for
2
2
2
the area of a triangle, we get
(A B C ) =

3(x + y + z)(2x + y)(2y + z)(2z + x)

...

16
4

Equation (2
...

Exercise 2
...
Let a, b and c be the lengths of the sides of a triangle, prove that
3(ab + bc + ca) ≤ (a + b + c)2 ≤ 4(ab + bc + ca)
...
19
...

Exercise 2
...
Let a, b and c be the lengths of the sides of a triangle, prove that
2 a2 + b2 + c2 ≤ (a + b + c)2
...
21
...

2
b+c c+a a+b
Exercise 2
...
(IMO, 1964) Let a, b and c be the lengths of the sides of a triangle,
prove that
a2 (b + c − a) + b2 (c + a − b) + c2 (a + b − c) ≤ 3abc
...
23
...

Exercise 2
...
(IMO, 1983) Let a, b and c be the lengths of the sides of a triangle,
prove that
a2 b(a − b) + b2 c(b − c) + c2 a(c − a) ≥ 0
...
25
...

a+b b+c c+a
8
Exercise 2
...
The lengths a, b and c of the sides of a triangle satisfy ab+bc+ca = 3
...

Exercise 2
...
Let a, b, c be the lengths of the sides of a triangle, and let r be the
inradius of the triangle
...

a b
c
2r
Exercise 2
...
Let a, b, c be the lengths of the sides of a triangle, and let s be the
semiperimeter of the triangle
...

4

Exercise 2
...
If a, b, c are the lengths of the sides of an acute triangle, prove that
a 2 + b 2 − c2

a 2 − b 2 + c2 ≤ a 2 + b 2 + c2 ,

cyclic

where

cyclic

stands for the sum over all cyclic permutations of (a, b, c)
...
30
...

2
...
3 The use of inequalities in the geometry of the
triangle
A problem which shows the use of inequalities in the geometry of the triangle was
introduced in the International Mathematical Olympiad in 1961; for this problem
there are several proofs and its applications are very broad, as will be seen later
on
...

Example 2
...
1
...

√

Since an equilateral triangle of side-length a has area equal to 43 a2 , the
equality in the example holds for this case; hence we will try to compare what
happens in any triangle with what happens in an equilateral triangle of side length
a
...
If AD is the altitude of the triangle at A, its length h can be expressed
=
as h = 23 a + y, where y measures its diﬀerence in comparison with the length
of the altitude of the equilateral triangle
...
We obtain
2
2
√
√ ah
a
a
a2 + b2 + c2 − 4 3(ABC) = a2 + h2 +
+ x + h2 +
−x −4 3
2
2
2
√
√
3
3
a+y
= a2 + 2h2 + 2x2 − 2 3 a
2
2
2
√
√
3
3 2
a + y + 2x2 − 3a2 − 2 3ay
= a +2
2
2
√
√
3
3
= a2 + a2 + 2 3ay + 2y 2 + 2x2 − 3a2 − 2 3ay
2
2
= 2(x2 + y 2 ) ≥ 0
...

60

Geometric Inequalities

Let us give another proof for the previous example
...
If we take d = AA , then d measures, in a manner,
how far is ABC from being an equilateral triangle
...

=
2
√
But d2 ≥ 0, hence we can deduce that 4 3(ABC) ≤ a2 + b2 + c2 , which is what
we wanted to prove
...

It is quite common to ﬁnd inequalities that involve elements of the triangle
among mathematical olympiad problems
...
36 of Section
1
...

(2
...

Another inequality, which has been very helpful to solve geometric-related
problems, is Nesbitt’s inequality (see Example 1
...
8 of Section 1
...
It states that
for a, b, c positive numbers, we always have
a
b
c
3
+
+
≥
...
4)

2
...
3) as follows:
b
c
a+b+c a+b+c a+b+c
a
+
+
=
+
+
−3
b+c c+a a+b
b+c
c+a
a+b
1
1
1
+
+
−3
= (a + b + c)
b+c c+a a+b
1
1
1
1
= [(a + b) + (b + c) + (c + a)] ·
+
+
−3
2
b+c c+a a+b
≥

9
3
−3=
...

Let us now observe some examples of geometric inequalities where such relationships are employed
...
3
...
Let ABC be an equilateral triangle of side length a, let M be a
point inside ABC and let D, E, F be the projections of M on the sides BC, CA
and AB, respectively
...

(ii)
MD + ME
ME + MF
MF + MD
a
A
F

B

M

E

D

C

Let x = M D, y = M E and z = M F
...

Therefore, h = x + y + z
...
8)
...
3) we can deduce that
h

1
1 1
+ +
x y z

≥ 9 and, after solving, that

√
1
1 1
9
6 3
+ + ≥ =

...
3), we can establish that
1
1
1
+
+
x+y y+z
z+x

(x + y + y + z + z + x)
Therefore,

1
x+y

+

1
y+z

+

1
z+x

9
2h

≥

=

≥ 9
...

Example 2
...
3
...

In order to prove the ﬁrst equation, observe that
larly,

r
hb

=

(ICA)
r
(ABC) , hc

=

(IAB)
(ABC)
...

Simi-

Adding the three equations, we have that

(ICA)
(IAB)
r
r
r
(IBC)
+
+
+
+
=
ha
hb
hc
(ABC) (ABC) (ABC)
(IBC) + (ICA) + (IAB)
= 1
...
3), since
1
1
1
(ha + hb + hc ) ha + hb + hc · r ≥ 9r
...
3
...
Let ABC be a triangle with altitudes AD, BE, CF and let H be
its orthocenter
...

HA
HB
HC
2

2
...
Since triangles ABC and HBC share the same base, their area ratio is
equal to their altitude ratio, that is, S1 = HD
...

S
AD
S
BE
S
CF
Then, HD + HE + HF = 1
...
3) we can state that
BE
CF
AD
+
+
HD HE
HF

HF
HD HE
+
+
AD
BE
CF

≥ 9
...

Moreover, the equality holds if and only if HD = HE = HF = 1 , that is,
AD
BE
CF
3
HD
if S1 = S2 = S3 = 1 S
...

Example 2
...
5
...
Let P , Q and R be the intersection points of
the lines AL, BM and CN with the circumcircle of ABC, respectively
...

LP
MQ NR
Let A be the midpoint of BC, let P be the midpoint of the arc BC, let D
and D be the projections of A and P on BC, respectively
...
Thus, the minimum value of AL + BM + CN
LP
LP
MQ
NR
is attained when P , Q and R are the midpoints of the arcs BC, CA and AB
...
Hence, without loss of generality, we will assume that AL, BM and CN
are the internal angle bisectors of ABC
...

a
b+c

2

...

a2

Similarly, for the internal angle bisectors BM and CN , we have
(c + a)2 − b2
BM
=
MQ
b2

and

(a + b)2 − c2
CN
=

...

3
≥

a+b
c

2

−3

2

−3

The ﬁrst inequality follows from the convexity of the function f (x) = x2 and the
b
second inequality from relations in the form a + a ≥ 2
...

Another way to ﬁnish the problem is the following:
2

2

2

b+c
c+a
a+b
+
+
−3
a
b
c
b2
c2
b2
c2
a2
a2
+ 2 +
+ 2 +
+ 2
=
2
2
2
b
a
c
b
a
c
≥ 2 · 3 + 2 · 3 − 3 = 9
...
3 The use of inequalities in the geometry of the triangle
Here we made use of the fact that
3

3

(ab)(bc)(ca)
a2 b2 c2

a2
b2

+

b2
a2

≥ 2 and that

65
ab
c2

+

bc
a2

+

ca
b2

≥

= 3
...
3
...
(Shortlist IMO, 1997) The lengths of the sides of the hexagon
ABCDEF satisfy AB = BC, CD = DE and EF = F A
...

BE
DA F C
2
B

A
a
C

c
b
F

D
E
Set a = AC, b = CE and c = EA
...
11), applied
to the quadrilateral ACEF , guarantees that AE · F C ≤ F A · CE + AC · EF
...
Therefore,
FA
c
≥

...

DE
DA

+

FA
FC

≥

a
b+c

+

and
b
c+a

+

c
a+b

DE
b
≥

...
31
...

a
b
c
Exercise 2
...
Let AD, BE, CF be the altitudes of the triangle ABC and let P Q,
P R, P S be the distances from a point P to the sides BC, CA, AB, respectively
...

PQ PR
PS
(i)

66

Geometric Inequalities

Exercise 2
...
Through a point O inside a triangle of area S three lines are drawn
in such a way that every side of the triangle intersects two of them
...
Prove that
1
1
9
1
+
+
≥ ,
(i)
S1
S2
S3
S
1
1
1
18
(ii)

...
34
...

BP
CP
Prove that AP + P M + P N = 6 if and only if P is the centroid of the triangle
...
35
...
Prove that
AD
BE
CF
(i)
+
+
≥ 9,
DD
EE
FF
BE
CE
9
AD
+
+
≥
...
36
...
Prove that
(i) la lb lc ≤ rs2 ,
(ii) la lb + lb lc + lc la ≤ s2 ,
2
2
2
(iii) la + lb + lc ≤ s2
...
37
...
Denote the lengths of the sides of the
triangle by a, b, c and the circumradius by R
...

AM
BN
CP
Exercise 2
...
Let ABC be a triangle with side-lengths a, b, c
...
Prove that
max {a ma , b mb , c mc } ≤ sR,
where R is the radius of the circumcircle and s is the semiperimeter
...
4 Euler’s inequality and some applications
Theorem 2
...
1 (Euler’s theorem)
...

2
...
Let us give a proof7 that depends only on Pythagoras theorem and the fact
that the circumcircle of the triangle BCI has center D, the midpoint of the arc8
BC
...

A

Q

I

O

B

M

C

D

Let M be the midpoint of BC and let Q be the orthogonal projection of I on the
radius OD
...

Therefore OI 2 = R2 − 2Rr
...

Theorem 2
...
2 (Euler’s inequality)
...
Moreover, R = 2r if and only if the
triangle is equilateral
...

proof can be found in [6, observation 3
...
7, page 123] or [1, page 76]
...

One of them is the following: the nine-point circle of a triangle is the circumcircle of the medial
triangle A B C
...
Clearly, a circle that intersects the three sides of a triangle
2
must have a greater radius than the radius of the incircle, therefore R ≥ r
...
4
...
In a triangle ABC, with circumradius R, inradius r and semiperimeter s, it happens that
s
R
r≤ √ ≤
...
We will use the fact that10 (ABC) = abc = sr
...
Thus,
√
8s3 ≥ 27(4Rrs) ≥ 27(8r2 s), since R ≥ 2r
...
√
s
The second inequality, 3√3 ≤ R , is equivalent to a + b + c ≤ 3 3R
...
Observe that the
last inequality holds because the function f√ = sin x is concave on [0, π], thus
(x)
sin A+sin B+sin C
≤ sin A+B+C = sin 60◦ = 23
...
39
...

Exercise 2
...
Let a, b and c be the lengths of the sides of a triangle, prove that
1
1
1
1
+
+
≥ 2,
ab bc ca
R
where R denotes the circumradius
...
41
...

sin A sin B
sin B sin C
sin C sin A
Exercise 2
...
Let A, B and C be the measurements of the angles in each of the
vertices of the triangle ABC, prove that
sin

A
2

sin

B
2

sin

C
2

≤

1

...
43
...
Call A, B and C the angles in the vertices
A, B and C, respectively
...
Prove that
2A
π

1
a

2B
π

1
b

2C
π

1
c

√

≤

2
3

3
R

...
4
...
In a triangle ABC with sides of length a, b
and c, and with circumcenter O, centroid G and circumradius R, the following
holds:
1 2
OG2 = R2 −
a + b 2 + c2
...

2
...
Let us use Stewart’s theorem which states11 that if L is a point on the side
BC of a triangle ABC and if AL = l, BL = m, LC = n, then a l2 + mn =
b2 m + c2 n
...

2
AA
3

and GA =

1
AA ,
3

substituting we get
2
2
1
OG2 + (A A)2 = A O2 · + R2 ·
...

=R −
9
= R2 −

One consequence of the last theorem is the following inequality
...
4
...
In a triangle ABC with side-lengths a, b and
c, with circumradius R, the following holds:
9R2 ≥ a2 + b2 + c2
...

[6, page 83] or [9, page 10]
...

Example 2
...
6
...

a+b+c

Using that 4R (ABC) = abc, we have the following equivalences:
3abc
a2 b 2 c2
a 2 + b 2 + c2
⇔ 4(ABC) ≤ √
≥

...

a+b+c

Exercise 2
...
Let A, B and C be the measurements of the angles in each of the
vertices of the triangle ABC, prove that
sin2 A + sin2 B + sin2 C ≤

9

...
45
...

Exercise 2
...
Suppose that the incircle of ABC is tangent to the sides BC, CA,
AB, at D, E, F , respectively
...

Exercise 2
...
Let a, b, c be the lenghts of the sides of a triangle ABC and let ha ,
hb , hc be the lenghts of the altitudes over BC, CA, AB, respectively
...

hb hc
hc ha
ha hb

2
...
The relationships that are most commonly used are
a + b + c = 2s,
ab + bc + ca = s2 + r2 + 4rR,

(2
...
6)

abc = 4Rrs
...
7)

2
...
Using Heron’s formula for the area of a triangle, we have
4R
the relationship s(s − a)(s − b)(s − c) = r2 s2 , hence
s3 − (a + b + c)s2 + (ab + bc + ca)s − abc = r2 s
...
5) and (2
...

Now, since any symmetric polynomial in a, b and c can be expressed as a polynomial in terms of (a + b + c), (ab + bc + ca) and (abc), it can also be expressed as a
polynomial in s, r and R
...

These transformations help to solve diﬀerent problems, as will be seen later
on
...
5
...
If A, B and C are the measurements of the angles within each of
r
the vertices of the triangle ABC, we have that cos A + cos B + cos C = R + 1
...

cos A + cos B + cos C =

c2 + a 2 − b 2
a 2 + b 2 − c2
b 2 + c2 − a 2
+
+
2bc
2ca
2ab

a b 2 + c 2 + b c 2 + a 2 + c a 2 + b 2 − a3 + b 3 + c 3
2abc
2
2
2
(a + b + c) (a + b + c ) − 2(a3 + b3 + c3 )
=
2abc
=

=

4s s2 − r2 − 4Rr − 4 s3 − 3r2 s − 6Rrs
8Rrs

s2 − r2 − 4Rr − (s2 − 3r2 − 6Rr)
2Rr
2
r
2r + 2Rr
=
+ 1
...
5
...
If A, B and C are the measurements of the angles in each of the
vertices of the triangle ABC, we have that cos A + cos B + cos C ≤ 3
...
5
...

r
R

+1, and using Euler’s

72

Geometric Inequalities
We can give another direct proof
...

Then,
b 2 + c2 − a 2
c2 + a 2 − b 2
a 2 + b 2 − c2
+
+
2bc
2ca
2ab
(b + c − a)(c + a − b)(a + b − c)
+ 1,
=
2abc

cos A + cos B + cos C =

and since (b + c − a)(c + a − b)(a + b − c) ≤ abc, we have the result
...
5
...
(IMO, 1991) Let ABC be a triangle, let I be its incenter and let
L, M , N be the intersections of the internal angle bisectors of A, B, C with BC,
AI BI CI
8
CA, AB, respectively
...

4
A
M
I
B

C
L
Using the angle bisector theorem BL = AB = c and the fact that BL +
LC
CA
b
ac
ab
LC = a, we can deduce that BL = b+c and LC = b+c
...
Hence,
AI + IL
IL
a
a+b+c
AL
=
=1+
=1+
=

...
13 Similarly, BM = a+b+c and CN =
inequality that we have to prove in terms of a, b and c is

a+b
a+b+c
...

≤
4
(a + b + c)3
27
The AM -GM inequality guarantees that
(b + c)(c + a)(a + b) ≤
13 Another

(b + c) + (c + a) + (a + b)
3

3

=

8
(a + b + c)3 ,
27

way to prove the identity is as follows
...
It is clear that AL = α+β+γ = r(a+c+b) = a+c+b
...
5 Symmetric functions of a, b and c

73

hence the inequality on the right-hand side is now evident
...

(a + b + c)3
(a + b + c)3
Substitute above, using equations (2
...
6) and (2
...

8s3
4
8s2
4
We can also use the Ravi transformation a = y + z, b = z + x, c = x + y, to reach
the ﬁnal result in the following way:
(b + c)(c + a)(a + b)
(x + y + z + x)(x + y + z + y)(x + y + z + z)
=
3
(a + b + c)
8(x + y + z)3
=

1
8

1+

x
x+y+z

=

1
8

1+

x + y + z xy + yz + zx
xyz
+
+
x+y+z
x+y+z
x+y+z

1+

y
x+y+z

1+

z
x+y+z
>

1

...
48
...

2
2
2
4

Exercise 2
...
Let a, b and c be the lengths of the sides of a triangle
...

a+b+c

Exercise 2
...
Let a, b and c be the lengths of the sides of a triangle
...

Exercise 2
...
(IMO, 1961) Let a, b and c be the lengths of the sides of a triangle,
prove that
√
4 3(ABC) ≤ a2 + b2 + c2
...
52
...

74

Geometric Inequalities

Exercise 2
...
Let a, b and c be the lengths of the sides of a triangle, prove that
√
4 3(ABC) ≤ ab + bc + ca
...
54
...

4 3(ABC) ≤
ab + bc + ca
Exercise 2
...
Let a, b and c be the lengths of the sides of a triangle
...

2
Exercise 2
...
Let a, b and c be the lengths of the sides of a triangle, let R and r
be the circumradius and the inradius, respectively, prove that
(b + c − a)(c + a − b)(a + b − c)
2r
=
...
57
...

b+c−a c+a−b a+b−c

Exercise 2
...
Let a, b and c be the lengths of the sides of a triangle
...
If τ1 = x + y + z, τ2 = xy + yz + zx and τ3 = xyz,
2
2
verify the following relationships
...

(2) a + b + c = 2τ1
...

2
(4) ab + bc + ca = τ1 + τ2
...

(6) 16(ABC)2 = 2(a2 b2 + b2 c2 + c2 a2 ) − (a4 + b4 + c4 ) = 16r2 s2 = 16τ1 τ3
...

√
4 τ1 τ3
τ3

...

2
...
6 Inequalities with areas and perimeters
We begin this section with the following example
...
6
...
(Austria–Poland, 1985) If ABCD is a convex quadrilateral of area
1, then
√
AB + BC + CD + DA + AC + BD ≥ 4 + 8
...
The area of
the quadrilateral ABCD is (ABCD) = ef sin θ , where θ is the angle between the
2
diagonals, which makes it clear that 1 = ef sin θ ≤ ef
...
Similarly, 1 = (ABCD) ≤ bc+da
...

Finally, since (e + f )2 = 4ef + (e − f )2 ≥ 4ef ≥ 8 and (a + b + c + d)2 =
4(a + c)(b + d) + ((a + c) − (b + d))2 ≥ 4(a + c)(b√ d) = 4(ab + bc + cd + da) ≥ 16,
+
we can deduce that a + b + c + d + e + f ≥ 4 + 8
...
6
...
(Iberoamerican, 1992) Using the triangle ABC, construct a hexagon H with vertices A1 , A2 , B1 , B2 , C1 , C2 as shown in the ﬁgure
...

A1

A2
a

a
A
c
B

b

B1

b
a

C c
c

b

C2
C1

B2
It is clear, using the area formula (ABC) =

ab sin C
,
2

that

(A1 A2 B1 B2 C1 C2 ) =(A1 BC2 ) + (A2 CB1 ) + (B2 AC1 ) + (AA1 A2 )
+ (BB1 B2 ) + (CC1 C2 ) − 2(ABC)
=

(c + a)2 sin B
(a + b)2 sin C
(b + c)2 sin A
+
+
2
2
2
+

c2 sin C
a2 sin A b2 sin B
+
+
− 2(ABC)
2
2
2

76

Geometric Inequalities
=

(a2 + b2 + c2 )(sin A + sin B + sin C)
+ ca sin B
2
+ ab sin C + b c sin A − 2(ABC)

=

(a2 + b2 + c2 )(sin A + sin B + sin C)
+ 4(ABC)
...

2
4R
sinA
1
a = 2R , we can
(a +b +c )(a+b+c)
≥ 9abc , that is,
4R
4R

Using the sine law,
if

2

2

prove that the inequality is true if and only

2

(a2 + b2 + c2 )(a + b + c) ≥ 9abc
...
Moreover, the equality holds only in the case a = b = c
...
6
...
(China, 1988 and 1993) Consider two concentric circles of radii R
and R1 (R1 > R) and a convex quadrilateral ABCD inscribed in the small circle
...
Show that
(i)

R1
perimeter of A1 B1 C1 D1
≥
;
perimeter of ABCD
R

(ii)

(A1 B1 C1 D1 )
≥
(ABCD)

2

R1
R

...
6 Inequalities with areas and perimeters

77

To prove (i), we use Ptolemy’s inequality (see Exercise 2
...
8)

DB1 · R1 ≤ A1 B1 · R + DA1 · R1
...

Therefore,
R1
perimeter (A1 B1 C1 D1 )
≥

...

A1

B1

x

A

w
d
D

a
O

c

B
y

b

z
C

D1

C1
◦

Since (AB1 C1 ) = x(a+y)sin2(180 −A) = x(a+y)sin A , we can produce the identity
2
(AB1 C1 )
x(a+y)
(BC1 D1 )
y(b+z) (CD1 A1 )
z(c+w) (DA1 B1 )
(ABCD) = ad+bc
...

Then,
x(a + y) + z(w + c) y(b + z) + w(d + x)
(A1 B1 C1 D1 )
=1+
+

...
In particular, the power of A1 , B1 , C1 and D1 is the same
...

78

Geometric Inequalities
Substituting this in the previous equation implies that the area ratio is

(A1 B1 C1 D1 )
z
y
w
x
2
= 1+(R1 −R2 )
+
+
+

...

(ad + bc)(ab + cd) ≤ ad+bc+ab+cd = (a+c)(b+d) ≤ 1 (a+b+c+d)2 ≤
4

= 8R2 , the ﬁrst two inequalities follow from the AM -GM inequality, and
the last one follows from the fact that, of all the quadrilaterals inscribed in a circle,
the square has the largest perimeter
...

Moreover, the equalities hold when ABCD is a square and only in this case
...
8) to identities, it must be the case that the four
quadrilaterals OAB1 C1 , OBC1 D1 , OCD1 A1 and ODA1 B1 are cyclic
...

There are problems that, even when they are not presented in a geometric form, they invite us to search for geometric relationships, as in the following
example
...
6
...
If a, b, c are positive numbers with c < a and c < b, we can deduce
√
that c(a − c) + c(b − c) ≤ ab
...

The area of the quadrilateral ABCD is, on the one hand,
(ABCD) = (ABC) + (ACD) =

c(a − c) +
√

b(b − c);

and on the other hand, (ABCD) = 2(ABD) = 2 ab sin ∠BAD
...

2
...
Since AC and BD are perpendiculars, Py√
√
thagoras theorem implies that DE = b√ c and EB = a − √ By √ √
−
Ptolemy’s
√
√
√ c
...
11), ( b − c + a − c) · (2 c) ≤ a b + a b and
then the result
...
59
...

The four points will form a quadrilateral of sides of length a, b, c and d, prove that
(i) 2 ≤ a2 + b2 + c2 + d2 ≤ 4,
√
(ii) 2 2 ≤ a + b + c + d ≤ 4
...
60
...
The six points form a hexagon of perimeter h
...

Exercise 2
...
Consider the three lines tangent to the incircle of a triangle ABC
which are parallel to the sides of the triangle; these, together with the sides of the
triangle, form a hexagon T
...

3

Exercise 2
...
Find the radius of the circle of maximum area that can be covered
using three circles with radius 1
...
63
...

Exercise 2
...
Two disjoint squares are located inside a square of side 1
...

Exercise 2
...
A convex quadrilateral is inscribed in a circumference of radius 1,
in such a way that one of its sides is a diameter and the other sides have lengths
a, b and c
...

80

Geometric Inequalities

Exercise 2
...
Let ABCDE be a convex pentagon such that the areas of the
triangles ABC, BCD, CDE, DEA and EAB are equal
...

2
(i)

Exercise 2
...
If AD, BE and CF are the altitudes of the triangle ABC, prove
that
perimeter (DEF ) ≤ s,
where s is the semiperimeter
...
68
...

Exercise 2
...
If a, b, c, d are the lengths of the sides of a convex quadrilateral,
show that
(i) (ABCD) ≤

ab + cd
,
2

(ii) (ABCD) ≤

ac + bd
and
2

(iii) (ABCD) ≤

a+c
2

b+d

...
7 Erd˝s-Mordell Theorem
o
Theorem 2
...
1 (Pappus’s theorem)
...
Let P be the intersection of B A with C A
...
Thus, we will have the following relationships between the areas:
(BP P C) = (AA B B) + (CC A A)
...
See the picture on the next page
...
7 Erd˝s-Mordell Theorem
o

81
P
A

A

A

B

C
B

C

P

P

Theorem 2
...
2 (Erd˝s-Mordell theorem)
...
If pa , pb , pc are the distances from P to the
sides of ABC, of lenghts a, b, c, respectively, then
P A + P B + P C ≥ 2 (pa + pb + pc )
...

Proof (Kazarinoﬀ)
...
Let A and C be the reﬂections of A and C
...

Now, let us consider the parallelograms determined by B, P and A , and by B, P
and C
...
The area of A P P C is at most b · P B
...
14 Then,
cpa + apc ≤ bP B
...

82

Geometric Inequalities
P

C

C

a
pc P

b

P

P

pa
c

B

A

A

B

Therefore,

a
c
P B ≥ pa + pc
...

a
a
c
c
If we add together these inequalities, we have
PA ≥

b c
+
c b

PA + PB + PC ≥

pa +

a
c
+
pb +
a
c

b
a
+
b
a

pc ≥ 2 (pa + pb + pc ) ,

b
since c + c ≥ 2
...

Example 2
...
3
...

Consider the two parallelograms that are determined by B, C, P and B,
A, P as shown in the ﬁgure, and the parallelogram that is constructed following
Pappus’s theorem
...

A

c

pc P

b

pa
B

a

C

2
...

Hence,
aP A + bP B + cP C ≥ 2(apa + bpb + cpc ) = 4(ABC)
...
7
...
Using the notation of the Erd˝s-Mordell theorem, prove that
o
pa P A + pb P B + pc P C ≥ 2 (pa pb + pb pc + pc pa )
...
Hence,
pa P A ≥

b
c
pa pb + pc pa
...

b
b
c
If we add together these three inequalities, we get

b
a
b c
+
pa pb +
+
b
a
c b
≥ 2 (pa pb + pb pc + pc pa )
...
7
...
Using the notation of the Erd˝s-Mordell theorem, prove that
o
2

1
1
1
+
+
PA PB
PC

≤

1
1
1
+
+
...
If A , B , C are
the inverse points of A, B, C, respectively, and A1 , B1 , C1 are the inverse points

84

Geometric Inequalities

of A1 , B1 , C1 , we can deduce that
P A · P A = P B · P B = P C · P C = d2 ,
P A1 · P A1 = P B1 · P B1 = P C1 · P C1 = d2
...

An application of the Erd˝s-Mordell theorem to the triangle A1 B1 C1 shows
o
that P A1 + P B1 + P C1 ≥ 2 (P A + P B + P C )
...

Example 2
...
6
...

2r
A

c

b

pc P
pa

B

c

C1

C

Let C1 be a point on BC such that BC1 = AB
...
Therefore,
2
PB ≥

pa + pc

...
7 Erd˝s-Mordell Theorem
o
Similarly,
PA ≥

85

pb + pc
2 sin A
2

and P C ≥

pa + pb

...

The solution of Exercise 2
...

sin B
2

sin C =
2

r
4R ,

Example 2
...
7
...
Prove that
at least one of the angles ∠P AB, ∠P BC, ∠P CA is less than or equal to 30◦
...
Using the Erd˝s-Mordell theorem we get P A + P B + P C ≥ 2P A1 +
o
2P B1 + 2P C1
...

If, for instance, P A ≥ 2P C1 , we can deduce that 1 ≥ P C1 = sin ∠P AB, then
2
PA
∠P AB ≤ 30◦ or ∠P AB ≥ 150◦
...

Example 2
...
8
...
Let RA , RC ,
RE denote the circumradii of triangles F AB, BCD, DEF , respectively, and let
P denote the perimeter of the hexagon
...

2

Let M , N and P be points inside the hexagon in such a way that M DEF ,
N F AB and P BCD are parallelograms
...
Observe that M N P and XY Z are similar
triangles
...
Similarly, Y N = 2RA and ZP = 2RC
...

The case M = N = P is the Erd˝s-Mordell inequality, on which the rest of the
o
proof is based
...
Let G, H denote the feet of the perpendiculars of M and X on Y Z ,
respectively
...

2
...

Since ∠XHG = 90◦ , then XH = XG sin ∠XGH ≤ XG
...

x
x

Similarly,
x
z
F N + BN ,
y
y
x
y
ZP ≥ BP + DP
...

x
x
y
y
z
z

(2
...

XY
YZ
ZX

If we apply the inequality

y
z

y
z
BP + BN =
z
y

+

z
y

≥ 2, we get

y
z
+
z
y

BP + BN
r
−
2
2
r yx zx
−

...

x
z
2 x
z
If we add the inequalities and substitute them in (2
...

...
70
...

r

Exercise 2
...
Using the notation of the Erd˝s-Mordell theorem, prove that
o
(i)

P B2
P C2
P A2
+
+
≥ 12,
pb pc
pc pa
pa pb

(ii)

PA
PB
PC
+
+
≥ 3,
pb + pc
pc + pa
pa + pb

PA
PB
PC
(iii) √
+√
+√
≥ 6,
pb pc
pc pa
pa pb
(iv) P A · P B + P B · P C + P C · P A ≥ 4(pa pb + pb pc + pc pa )
...
72
...
If, for example, P and A are on diﬀerent sides of the segment BC,
then pa is negative, and we have a similar situation for the other cases
...

2
...

The Fermat-Steiner problem
...
We will present three solutions and point out
the methods used to solve the problem
...
It takes as its starting point the following two lemmas
...
8
...
The sum of the distances from an interior point
to the sides of an equilateral triangle is equal to the altitude of the triangle
...
Let P be a point in the interior of the triangle ABC
...

2
...
Moreover, P M is
the altitude of the equilateral triangle AP P
...
Let L be a point on B C such that
A L is the altitude of the triangle A B C from A
...

Next, we present another two proofs of Viviani’s lemma for the sake of completeness
...
8
...
(i) The following is another proof of Viviani’s lemma which
is based on the use of areas
...
Then, if a is the length of the side of the triangle and h is the
length of its altitude, we have that ah = aP N + aP L + aP M , that is, h =
PN + PL + PM
...

A

N

M
P
M

B

L

C

90

Geometric Inequalities

Lemma 2
...
3
...
The point P is
known as the Fermat point of the triangle
...
First, we will proof the existence of P
...
Their circumcircles intersect at A
and at another point that we denote as P
...
Similarly, since
AP BC is cyclic, ∠AP B = 120◦
...

To prove the uniqueness, suppose that Q satisﬁes ∠AQB = ∠BQC =
∠CQA = 120◦
...
Similarly, it should be on the circumcircle of CAB , hence Q = P
...
Given
the triangle ABC with angles less than or equal to 120◦ , construct the Fermat
point P , which satisﬁes ∠AP B = ∠BP C = ∠CP A = 120◦
...

These perpendiculars determine a triangle DEF which is equilateral
...

Now, since ∠BP C = 120◦, we can deduce that ∠BDC = 60◦
...
Therefore DEF is indeed equilateral
...
Observe that
any other point Q inside the triangle ABC satisﬁes AQ ≥ A Q, where A Q is the
distance from Q to the side EF , similarly BQ ≥ B Q and CQ ≥ C Q
...

2
...
This way of solving the problem uses the ingenious idea
of rotating the ﬁgure to place the three segments that we need next to each other,
in order to form a polygonal line and then add them together
...
This proof was provided by J
...
Let us recall this solution
...

Next, rotate the ﬁgure with its center in B and through an angle of 60◦ , in a
positive direction
...
If C is the image of A and P is the image
of P , the triangles BP P and BAC are equilateral
...
The path
CP + P P + P C is minimum when C, P , P and C are collinear, which in turn
requires that ∠C P B = 120◦ and ∠BP C = 120◦; but since ∠C P B = ∠AP B,

92

Geometric Inequalities

the point P should satisfy ∠AP B = ∠BP C = 120◦ (and then also ∠CP A = 120◦)
...
If we review the proof, we can see that
the point P is on the segment CC , where C is the third vertex of the equilateral
triangle with side AB
...
Hence we can ﬁnd P as the intersection of BB and
CC
...
When we solve maximum and minimum problems we are principally faced with three questions, (i) is there a solution?, (ii) is there a unique
solution? (iii) what properties characterize the solution(s)? Torricelli’s solution
demonstrates that among all the points in the triangle, this particular point P ,
from which the three sides of the triangle are observed as having an angle of 120◦ ,
provides the minimum value of P A + P B + P C
...
However, the solution does not give us any clue as to why Torricelli chose this point, or what made
him choose that point; probably this will never be known
...
These ideas belong to the Swiss geometer Jacob Steiner
...

Lemma 2
...
4 (Heron’s problem)
...

B
A
d
P
The shortest path between A and B, touching the line d, can be found reﬂecting B on d to get a point B ; the segment AB intersects d at a point P ∗ that
makes AP ∗ + P ∗ B represent the minimum between the numbers AP + P B, with
P on d
...

This point satisﬁes the following reﬂection principle: The incident angle is
equal to the reﬂection angle
...

2
...
8
...
Given two points A and B outside
the circle C, ﬁnd the shortest path that starts at A, touches the circle and ﬁnishes
at B
...

Let D be a point on C, then we have that the set {P : P A+P B = DA+DB}
is an ellipse ED with foci points A and B, and that the point D belongs to ED
...
Moreover, these ellipses have the property that Ed
is a subset of the interior of Ed if and only if d < d
...

The optimal point Q will belong to an ellipse, precisely to EQ
...

Thus, the point Q that minimize AQ + QB should satisfy that the ellipse EQ
is tangent to C
...

C

α
A

Q

β

B

Now let us go back to Steiner’s solution of the Fermat-Steiner problem
...
In the ﬁrst case, if
P is one of the vertices, then one term of the sum P A + P B + P C is zero and the
other two are the lengths of the sides of the triangle ABC that have in common
the chosen vertex
...

In order to analize the second case, Steiner follows the next idea (very useful
in optimization problems and one which can be taken to belong to the strategy of
“divide and conquer”), which is to keep ﬁxed some of the variables and optimize
the rest
...
Such
restrictions will act as restrains in the solution space until we reach the optimal
solution
...
Suppose that P A is ﬁxed; that is, P
belongs to the circle of center A and radius P A, where we need to ﬁnd the point
P that makes the sum P B + P C minimum
...
From this, it follows that P A+P B = P C > AB +BC, which means B would
be a more suitable point (instead of P )
...
Now, since B and C are points outside the circle C = (A, P A), the optimal
point for the problem of minimizing P B + P C with the condition that P is on
the circle C is, by Lemma 2
...
5, a point Q on the circle C, such that this circle is
tangent to the ellipse with foci B and C in Q, and the point Q is such that the
angles ∠AQB and ∠CQA are equal
...
This means
Q should be the Fermat point
...

The Fagnano problem
...
We present two classical solutions, where the
reﬂection on lines play a central role
...
Schwarz and the other to
L
...

2
...
The German mathematician Hermann Schwarz provided the
following solution to this problem for which he took as starting point two observations that we present as lemmas
...
Such a triangle is known as the ortic triangle
...
8
...
Let ABC be a triangle, and let D, E and F be the feet of the altitudes
on BC, CA and AB as they fall from the vertices A, B and C, respectively
...

Proof
...

A

E
F
H

B

C
D

Since these two triangles have a common angle at A, it is suﬃcient to see that
∠AEF = ∠ABC
...

Lemma 2
...
7
...

Proof
...

A
D
E
F
D
B

C
D

96

Geometric Inequalities

Using these elements we can now continue with the solution proposed by H
...

We will now prove that the triangle with minimum perimeter is the ortic triangle
...

C
E

M

B

L

A
N

N
F

D

F

B

A

A

B

C
Reﬂect the complete ﬁgure on the side BC, so that the resultant triangle is reﬂected on CA, then on AB, on BC and ﬁnally on CA
...
The side AB of the last triangle
is parallel to the side AB of the ﬁrst, since as a result of the ﬁrst reﬂection, the
side AB is rotated in a negative direction through an angle of 2B, and then in a
negative direction through an angle of 2A, the third reﬂection is invariant and the
fourth is rotated through an angle of 2B in a positive direction and in the ﬁfth it
is also rotated in a positive direction through an angle of 2A
...

The segment F F is twice the perimeter of the ortic triangle, since F F is
composed of six pieces where each side of the ortic triangle is taken twice
...
Moreover, N N is parallel to
the line F F and of the same length, then since the length of the broken line N N
is greater than the length of the segment N N , we can deduce that the perimeter
of DEF is less than the perimeter of LM N
...
The solution due to the Hungarian mathematician L
...
Let LM N be a triangle inscribed on ABC
...
It is clear that LM = M L and
L N = N L
...

2
...
Now, let us see which is the best option for the point L
...

A

M

N

L

L
B

L

C

It is evident that AL = AL = AL and that AC and AB are internal angle
bisectors of the angles LAL and L AL, respectively
...
The cosine law applied to the triangle AL L guarantees
that
(L L )2 = (AL )2 + (AL )2 − 2AL · AL cos 2α
= 2AL2 (1 − cos 2α)
...
15 A similar analysis using the points B and C will demonstrate that
15 This would be enough to ﬁnish Fejer’s proof for the Fagnano’s problem
...
Let us see why
...
Therefore ∠CLE = ∠A
...
Similarly, it follows that CE is an altitude
...
Thus, the triangle LM N with minimum
perimeter is the ortic triangle
...
73
...
If O is the intersection
of the diagonals AC and BD, and P , Q, R, S are the feet of the perpendiculars of O
on the sides AB, BC, CD, DA, respectively, prove that P QRS is the quadrilateral
of minimum perimeter inscribed in ABCD
...
74
...
Let D, E and F be
the points of intersection of AP , BP and CP with the sides BC, CA and AB,
respectively
...

Exercise 2
...
(IMO, 1981) Let P be a point inside the triangle ABC
...

BC
CA
Find the point P that minimizes P D + P E + AB
...
76
...
75
...
77
...
75
...
78
...
For which point P is the
sum of P A2 + P B 2 + P C 2 minimum?
Exercise 2
...
For every point P on the circumcircle of a triangle ABC, we draw
the perpendiculars P M and P N to the sides AB and CA, respectively
...

Exercise 2
...
(Turkey, 2000) Let ABC be an acute triangle with circumradius R;
let ha , hb and hc be the lengths of the altitudes AD, BE and CF , respectively
...
Prove that
t2
t2
t2
3
a
+ b + c ≤ R
...
81
...
Prove that
(i)

ha
hb
hc
+
+
≥ 9,
pa
pb
pc

(ii) ha hb hc ≥ 27pa pb pc ,
(iii) (ha − pa )(hb − pb )(hc − pc ) ≥ 8pa pb pc
...
82
...

2
...
83
...

Exercise 2
...
Of all triangles with a given perimeter, the one with largest area is
the equilateral triangle
...
85
...

Exercise 2
...
If P is a point inside the triangle ABC, l = P A, m = P B and
n = P C, prove that
(lm + mn + nl)(l + m + n) ≥ a2 l + b2 m + c2 n
...
87
...

Exercise 2
...
Let (ABC) be the area of a triangle ABC and let F be the Fermat
point of the triangle
...

Exercise 2
...
Let P be a point inside the triangle ABC, prove that
P A + P B + P C ≥ 6r
...
90
...
For a triangle ABC and a point
P on the plane, we deﬁne the “pedal triangle” of P with respect to ABC as the
triangle A1 B1 C1 where A1 , B1 , C1 are the feet of the perpendiculars from P to
BC, CA, AB, respectively
...
We can thus conclude that the pedal triangle of
maximum area is the medial triangle
...
1
...
Suppose that A1 A2 = B1 B3 = C1 C4 , prove that
√
S B + SC
1
<
< 2 − 2
...
2
...

Prove that CD ≥ AB
...
3
...
Prove that if the inradius of
the triangle is r, then
a sin α + b sin β + c sin γ ≥ 9r
...
4
...
Let S be the set of points which are interior to at least two of the
circles
...
5
...

If S = area(CDE) and r is the inradius of ABC, prove that S ≥ 2r2
...
6
...
Let B be a point such√
that ∠ABX = 120◦ and let C be a point between
B and X
...

102

Recent Inequality Problems

Problem 3
...
(Korea, 1995) A ﬁnite number of points on the plane have the property that any three of them form a triangle with area at most 1
...

Problem 3
...
(Poland, 1995) For a ﬁxed positive integer n, ﬁnd the minimum
value of the sum
x3
xn
x2
x1 + 2 + 3 + · · · + n ,
2
3
n
given that x1 , x2 ,
...

Problem 3
...
(IMO, 1995) Let ABCDEF be a convex hexagon with AB = BC =
CD and DE = EF = F A such that ∠BCD = ∠EF A = π
...
Prove that
3
AG + GB + GH + DH + HE ≥ CF
...
10
...
Let R and r be the circumradius and the inradius of the triangle
...

Problem 3
...
(China, 1996) Suppose that x0 = 0, xi > 0 for i = 1, 2,
...
Prove that
n

√

1≤
i=1

xi
π
√
<
...
12
...
, an ∈ R+ with
n
Prove that for x1 , x2 ,
...

ai x2
i

...
13
...
, xn , xn+1 be positive real numbers
with x1 + x2 + · · · + xn = xn+1
...

i=1

Problem 3
...
(St
...
Prove that OM ≥ ON
...
15
...

Problem 3
...
(Taiwan, 1997) Let a1 ,
...
, n, a0 = an , an+1 = a1 and n ≥ 3
...

a1
a2
a3
an

Problem 3
...
(Taiwan, 1997) Let ABC be an acute triangle with circumcenter
O and circumradius R
...

Problem 3
...
(APMO, 1997) Let ABC be a triangle
...
Let la = AX
...
Prove that
la
lb
lc
+
+
≥3
2
2
sin A sin B
sin2 C
with equality if and only if the triangle is equilateral
...
19
...
, xn be real numbers satisfying
|x1 + · · · + xn | = 1 and |xi | ≤ n+1 for all i = 1,
...
Prove that there exists
2
a permutation y1 ,
...
, xn such that
|y1 + 2y2 + · · · + nyn | ≤

n+1

...
20
...
A triangle exists with sides of lengths a, b and c if and only if there exist
numbers x, y and z such that
y
z
a
+ = ,
z
y
x

z
x
b
+ = ,
x z
y

x y
c
+ =
...
21
...
Prove that the
area of the triangle P QR is at most one-half of the area of the hexagon
...
22
...
Prove that
x3 + x3 + x3 + x3 ≥ max x1 + x2 + x3 + x4 ,
1
2
3
4

1
1
1
1
+
+
+
x1
x2
x3
x4

...
23
...
Prove that
√
√
x+y+z ≥ x−1+

y−1+

√
z − 1
...
24
...

If M is a point on the arc AB, prove that
√
M C · M D ≥ 3 3M A · M B
...
25
...
If the lines AP , BP and CP intersect the sides BC, CA and AB
of the triangle at L, M and N , respectively, prove that
P L + P M + P N < a
...
26
...
Prove that
P B QC
1
·
≤
...
27
...
The lines CO, AO and BO intersect the circumcircles of
the triangles AOB, BOC and AOC, for the second time, at C1 , A1 and B1 ,
respectively
...

OA1
OB1
OC1
2
Problem 3
...
(Balkan, 1999) Let ABC be an acute triangle and let L, M , N be
the feet of the perpendiculars from the centroid G of ABC to the sides BC, CA,
AB, respectively
...

27
(ABC)
4
Problem 3
...
(Belarus, 1999) Let a, b, c be positive real numbers such that
a2 + b2 + c2 = 3
...

1 + ab 1 + bc 1 + ca
2
Problem 3
...
(Czech and Slovak Republics, 1999) For arbitrary positive numbers
a, b and c, prove that
a
b
c
+
+
≥ 1
...
31
...
Prove that
b2
c2
d2
1
a2
+
+
+
≥
...
32
...
Prove that
DE ≤

AB + BC + CA

...
33
...
The point E on CA is deﬁned by the equation EC = AD−BC
...

Problem 3
...
(Romania, 1999) Let a, b, c be positive real numbers such that
ab + bc + ca ≤ 3abc
...

Problem 3
...
(Romania, 1999) Let x1 , x2 ,
...
Prove that
1
1
1
+
+ ··· +
≤ 1
...
36
...
, xn , yn be positive real numbers such that x1 + x2 + · · · + xn ≥ x1 y1 + x2 y2 +
· · · + xn yn
...

y1
y2
yn

Problem 3
...
(Russia, 1999) Let a, b and c be positive real numbers with abc = 1
...

Problem 3
...
(Russia, 1999) Let {x} = x − [x] denote the fractional part of x
...

2

Problem 3
...
(Russia, 1999) The positive real numbers x and y satisfy x2 + y 3 ≥
x3 + y 4
...

106

Recent Inequality Problems

Problem 3
...
(St
...

Prove that
x0 +

1
1
1
+
+ ···+
≥ xn + 2n
...
41
...

Problem 3
...
(United Kingdom, 1999) Three non-negative real numbers a, b and
c satisfy a + b + c = 1
...

Problem 3
...
(USA, 1999) Let ABCD be a convex cyclic quadrilateral
...

Problem 3
...
(APMO, 1999) Let {an } be a sequence of real numbers satisfying
ai+j ≤ ai + aj for all i, j = 1, 2,
...

Problem 3
...
(IMO, 1999) Let n ≥ 2 be a ﬁxed integer
...
, xn
...

Problem 3
...
(Czech and Slovak Republics, 2000) Prove that for all positive real
numbers a and b,
3

a
+
b

3

b
≤
a

3

2(a + b)

1 1
+

...
47
...
Prove that
a2 x2
b2 y 2
c2 z 2
3
+
+
≥
...
48
...
Prove that
4(AP 2 + BQ2 + CR2 + DS 2 ) ≤ 5(AB 2 + BC 2 + CD2 + DA2 )
...
49
...
Prove that
2 ≤ (1 − x2 )2 + (1 − y 2 )2 + (1 − z 2 )2 ≤ (1 + x)(1 + y)(1 + z)
...
50
...

Prove that
1
1
1
b−1+
c−1+
≤ 1
...
51
...
Prove that
√
a2 + b2 + c2 ≥ 3 abc
...
52
...

Problem 3
...
(Poland, 2001) Prove that the inequality
n

ixi ≤
i=1

n
2

n

xi
i

+
i=1

holds for every integer n ≥ 2 and for all non-negative real numbers x1 , x2 ,
...

Problem 3
...
(Austria–Poland, 2001) Prove that
2<

a + b b + c c + a a 3 + b 3 + c3
+
+
−
≤ 3,
c
a
b
abc

where a, b, c are the lengths of the sides of a triangle
...
55
...

2 + 8bc
2 + 8ca
2 + 8ab
a
b
c
Problem 3
...
(Short list IMO, 2001) Let x1 , x2 ,
...

2 + 1 + x2 + x2 + · · · + 1 + x2 + · · · + x2 <
1 + x1
n
1
2
1
Problem 3
...
(Austria, 2002) Let a, b, c be real numbers such that there exist
α, β, γ ∈ {−1, 1} with αa + βb + γc = 0
...

Problem 3
...
(Balkan, 2002) Prove that
2
2
27
2
+
+
≥
b(a + b) c(b + c) a(c + a)
(a + b + c)2
for positive real numbers a, b, c
...
59
...

Problem 3
...
(Ireland, 2002) Prove that
√
3 3 xyz
x
y
z
+
+
≥
√
1−x 1−y 1−z
1 − 3 xyz
for positive real numbers x, y, z less than 1
...
61
...
Prove that
1
a
+
b+c 2

b
1
+
c+a 2

c
1
+
a+b 2

≥ 1
...
62
...
Prove that
1 1 1
a+b b+c c+a
9
+ + +
...
63
...

Problem 3
...
(APMO, 2002) Let a, b, c be positive real numbers satisfying
1
1
b + c = 1
...

Problem 3
...
(Ireland, 2003) The lengths a, b, c of the sides of a triangle are such
that a + b + c = 2
...

27

Problem 3
...
(Romania, 2003) Prove that in any triangle ABC the following
inequality holds:
√
1
3
1
1
,
+
+
≤
mb mc
mc ma
ma mb
S
where S is the area of the triangle and ma , mb , mc are the lengths of the medians
...
67
...
Prove that
1 + ab 1 + bc 1 + cd 1 + da
+
+
+
≥ 4
...
68
...
Prove that
√
la + lb + lc ≤ 3s
...
69
...
Prove that
1
1
2
2
2
1
+
+
≥
+
+

...
70
...

Problem 3
...
(IMO, 2003) Given n > 2 and real numbers x1 ≤ x2 ≤ · · · ≤ xn ,
prove that
⎛
⎞2
2
⎝
|xi − xj |⎠ ≤ (n2 − 1)
(xi − xj )2 ,
3
i,j
i,j
where equality holds if and only if x1 , x2 ,
...

Problem 3
...
(Short list Iberoamerican, 2004) If the positive numbers x1 , x2 ,
...

x2 (x1 + x2 + x3 ) x3 (x2 + x3 + x4 )
x1 (xn + x1 + x2 )
3
Problem 3
...
(Czech and Slovak Republics, 2004) Let P (x) = ax2 + bx + c be
a quadratic polynomial with non-negative real coeﬃcients
...

x
Problem 3
...
(Croatia, 2004) Prove that the inequality
a2
b2
c2
3
+
+
≥
(a + b)(a + c) (b + c)(b + a) (c + a)(c + b)
4
holds for all positive real numbers a, b, c
...
75
...
Prove that
1
1
1
+
+
≥ 1
...
76
...

x4 + y 4 + z 4 + 3(x + y + z) ≥
y
z
x
z
x
y
Problem 3
...
(Korea, 2004) Let R and r be the circumradius and the inradius
of the acute triangle ABC, respectively
...
Let M be the midpoint of BC and let X be the intersection of the tangents
to the circumcircle of ABC at B and C
...

R
AX
Problem 3
...
(Moldova, 2004) Prove that for all real numbers a, b, c ≥ 0, the
following inequality holds:
√
√
√
a3 + b3 + c3 ≥ a2 bc + b2 ca + c2 ab
...
79
...
Prove that
√
√
√
√
√
√
xy + z + yz + x + zx + y ≥ 1 + xy + yz + zx
...
80
...
Prove that
a3 + b3 + c3 ≥ ab + bc + ca
...
81
...

Problem 3
...
(Romania, 2004) The real numbers a, b, c satisfy a2 + b2 + c2 = 3
...

Problem 3
...
(Romania, 2004) Consider the triangle ABC and let O be a point in
the interior of ABC
...
Let R1 , R2 , R3 be the radii of the circumcircles of
the triangles OBC, OCA, OAB, respectively, and let R be the radius of the
circumcircle of the triangle ABC
...

AA1
BB1
CC1
Problem 3
...
(Romania, 2004) Let n ≥ 2 be an integer and let a1 , a2 ,
...
Prove that for any non-empty subset S ⊂ {1, 2,
...

≤
1≤i≤j≤n

Recent Inequality Problems

111

Problem 3
...
(APMO, 2004) For any positive real numbers a, b, c, prove that
(a2 + 2)(b2 + 2)(c2 + 2) ≥ 9(ab + bc + ca)
...
86
...
Prove that
√
3
1
3
3
+ 6(a + b + c) ≤

...
87
...
Let t1 , t2 ,
...

Prove that ti , tj , tk are the side-lengths of a triangle for all i, j, k with 1 ≤ i <
j < k ≤ n
...
88
...
Prove that
√
√
√
3
3
a 1 + b − c + b 3 1 + c − a + a 1 + a − b ≤ 1
...
89
...
, x6 be real numbers such that x2 +
1
x2 + · · · + x2 = 6 and x1 + x2 + · · · + x6 = 0
...

2
6
2
Problem 3
...
(United Kingdom, 2005) Let a, b, c be positive real numbers
...

≥ (a + b + c)
b
c a
a b
c
Problem 3
...
(APMO, 2005) Let a, b and c be positive real numbers such that
abc = 8
...

3

Problem 3
...
(IMO, 2005) Let x, y, z be positive real numbers such that xyz ≥ 1
...

5 + y2 + z 2
2 + x2
x
y +z
z + x2 + y 2
Problem 3
...
(Balkan, 2006) Let a, b, c be positive real numbers, prove that
1
1
1
3
+
+
≥

...
94
...
Prove that the area of the triangle ABC is less than or equal
to the area of the triangle A B C
...
95
...

Problem 3
...
(Turkey, 2006) Let a1 , a2 ,
...

1
2
n
Prove that
i=j

(n − 1)2 A
ai
≥

...
97
...
, an ,
not necessarily distinct
...
Prove that
(n − 1)d ≤ s ≤

n2 d
,
4

and determine the conditions on the n numbers that ensure the validity of the
equalities
...
98
...

Problem 3
...
(Bulgaria, 2007) Find all positive integers n such that if a, b, c are
non-negative real numbers with a + b + c = 3, then
abc(an + bn + cn ) ≤ 3
...
100
...

3 2 2
3
3
3 c a +1
3 a2 b 2 + 1
3 b 2 c2 + 1
Problem 3
...
(China, 2007) If a, b, c are the lengths of the sides of a triangle
with a + b + c = 3, ﬁnd the minimum of
a2 + b2 + c2 +

4abc

...
102
...

b(b + c − a) c(c + a − b) a(a + b − c)
Problem 3
...
(Iran, 2007) If a, b, c are three diﬀerent positive real numbers,
prove that
a+b b+c c+a
+
+
> 1
...
104
...
Prove that xz < 1
...
105
...
Prove that
[x]
x + {x}
−
[x]
x + {x}

+

{x}
x + [x]
−
{x}
x + [x]

>

9
,
2

where [x] and {x} represent the integer part and the fractional part of x, respectively
...
106
...
Prove that
b
c
a+b+c≥

2
3
+

...
107
...

a+b+1 b+c+1 c+a+1
Prove that
a + b + c ≥ ab + bc + ca
...
108
...

For every interior point P of ABC, consider the circle with center A and radius
AP ; let M and N be the intersections of the sides AB and AC with the circle,
respectively
...

Problem 3
...
(Romania, 2007) The points M , N , P on the sides BC, CA, AB,
respectively, are such that the triangle M N P is acute
...
Prove that x ≤ 2X
...
110
...
Prove that
x2 + yz
+
2x2 (y + z)

y 2 + zx
+
2y 2 (z + x)

z 2 + xy
≥ 1
...
111
...

+
+
≥
2 + b + c2
2 + c + a2
2 + a + b2
12
Under which circumstances the equality holds?
Problem 3
...
(Canada, 2008) Let a, b, c be positive real numbers for which
a + b + c = 1
...

a + bc b + ca c + ab
2
Problem 3
...
(Iran, 2008) Find the least real number K such that for any
positive real numbers x, y, z, the following inequality holds:
√
√
√
x y + y z + z x ≤ K (x + y)(y + z)(z + x)
...
114
...

32

Problem 3
...
(Ireland, 2008) Let x, y, z be positive real numbers such that
xyz ≥ 1
...

The equalities hold if and only if x = y = z = 1
...
116
...

1 + a2 (b + c) 1 + b2 (c + a) 1 + c2 (a + b)
abc
Problem 3
...
(Romania, 2008) Determine the maximum value for the real number k if
1
1
1
(a + b + c)
+
+
−k ≥k
a+b b+c c+a
for all real numbers a, b, c ≥ 0 and with a + b + c = ab + bc + ca
...
118
...
Prove that
a2 + b2 + c2 + 3abc ≥

4

...
119
...

Prove that
1
1
1
4

...
120
...

2
2
(x − 1)
(y − 1)
(z − 1)2
(ii) Prove that the equality holds for an inﬁnite number of x, y, z, all of them
being rational numbers
...
In Sections 1 and 2 we provide the solutions to the exercises
in Chapters 1 and 2, respectively, and in Section 3 the solutions to the problems in
Chapter 3
...

4
...
1
...
1
...

Solution 1
...
(i) If a < 0, then −a > 0
...

(ii) (−a)b > 0
...
1
...

(iv) Use property 1
...
2
...

1
(vi) a a = 1 > 0
...

(viii) Use (vi) and property 1
...
3
...

(x) Use property 1
...
3 with a − 1 > 0 and a > 0
...
1
...

Solution 1
...
(i) a2 < b2 ⇔ b2 − a2 = (b + a)(b − a) > 0
...
1
...

b
Solution 1
...
For (i), (ii) and (iii) use the deﬁnition, and for (iv) and (v) remember
that |a|2 = a2
...
5
...

(ii) Consider |a| = |a − b + b| and |b| = |b − a + a|, and apply the triangle inequality
...

(iv) (x2 − xy + y 2 )(x + y) = x3 + y 3
...
6
...
Then, we can assume |a| ≥
|b| ≥ |c| > 0
...

a
a
a
a a
a
a a

b
c
Since a ≤ 1 and a ≤ 1, we can deduce that 1 +
Thus, it is suﬃcient to prove that

b
c
b
c
b
c
+
−
+
− 1+ +
a
a
a a
a a

b
a

b
= 1+ a and 1 +

+ 1+

c
a

c
= 1+ a
...

a a

Now, use the triangle inequality and Exercise 1
...

Solution 1
...
(i) Use that 0 ≤ b ≤ 1 and 1 + a > 0 in order to see that
0 ≤ b(1 + a) ≤ 1 + a ⇒ 0 ≤ b − a ≤ 1 − ab ⇒ 0 ≤

b−a
≤ 1
...
Since 1 + a ≤ 1 + b, it follows
1
1
that 1+b ≤ 1+a , and then prove that
a
b
a
b
a+b
+
≤
+
=
≤ 1
...
For the inequality on the right-hand side,
note that b ≤ 1 ⇒ b2 ≤ b ⇒ −b ≤ −b2 , and then
ab2 − ba2 ≤ ab2 − b2 a2 = b2 (a − a2 ) ≤ a − a2 =
Solution 1
...
Prove in general that x <
√
1
1 + 1+x < 2
...

4
2
4
√
√
> 2 and that x > 2 ⇒

Solution 1
...
ax + by ≥ ay + bx ⇔ (a − b)(x − y) ≥ 0
...
10
...
Then, use the previous exercise substi√
1
1
tuting with x2 , y 2 , √y and √x
...
11
...

4
...
12
...

Solution 1
...
In order for the expressions in the inequality to be well deﬁned, it
1
is necessary that x ≥ − 2 and x = 0
...
Perform some simpliﬁcations and show that 2 2x + 1 < 7;
then solve for x
...
14
...
1 Hence, its integer part is 2n and then we have to prove that
4n2 + n < 2n + 4 , this follows immediately after squaring both sides of the
inequality
...
15
...

=
5 + b5 + ab
2
a
a b (a + b) + ab
a b c (a + b) + abc
a+b+c
Similarly,

bc
b5 +c5 +bc

≤

a
a+b+c

and

ca
c5 +a5 +ca

≤

b
a+b+c
...

Solution 1
...
Consider p(x) = ax2 + bx + c, using the hypothesis, p(1) = a + b + c
and p(−1) = a − b + c are not negative
...
If x1 , x2 are the roots of p, we can
2a
4a
b
c
deduce that a = −(x1 + x2 ) and a = x1 x2 , therefore a+b+c = (1 − x1 )(1 − x2 ),
a
a−b+c
a−c
= (1 + x1 )(1 + x2 ) and a = 1 − x1 x2
...

Solution 1
...
If the inequalities are true, then a, b and c are less than 1, and
1
a(1 − b)b(1 − c)c(1 − a) > 64
...

Solution 1
...
Use the AM-GM inequality with a = 1, b = x
...
19
...

Solution 1
...
Use the AM-GM inequality with a = x2 , b = y 2
...
21
...

Solution 1
...
Use the AM-GM inequality with a = x+y , b = x+y and also use
x
y
the AM-GM inequality for x and y
...
20
...
23
...
24
...
25
...

a
b

b
x
...

√
√
√
( a− b)2
− ab =
, simplify and ﬁnd the bounds using 0 < b ≤
2

a+b
2

√
Solution 1
...
x + y ≥ 2 xy
...
27
...

√
Solution 1
...
xy + zx ≥ 2x yz
...
29
...
27
...
30
...
31
...
32
...

xy

≥
yz
x

xy 2 z
zx

≥2

= 2y
...

≥x

Solution 1
...
x4 + y 4 + 8 = x4 + y 4 + 4 + 4 ≥ 4 4 x4 y 4 16 = 8xy
...
34
...

b
c
Solution 1
...

a
b

+

b
c

+

c
d

+

d
a

≥44

abc d
b cda

= 4
...
36
...
37
...
38
...
39
...

= n
...

Solution 1
...
Using the AM-GM inequality, we obtain
3 b3
b3
a3
3 a
+
+ bc ≥ 3
·
· bc = 3ab
...

n−1
2

(a −

4
...

c3
a

121

3

3

3

3

+ ab + ab ≥ 3ca
...
The inequality can also be proved using Exercise 1
...

Solution 1
...
If abc = 0, the result is clear
...

Solution 1
...
Apply the AM-GM inequality twice over, a2 b + b2 c + c2 a ≥ 3abc,
ab2 + bc2 + ca2 ≥ 3abc
...
43
...
44
...

(a + b + c) ≥

1
1
1
+
+
a+b b+c c+a

1+a
1+b

+ bc

9
2

is equivalent to

(a + b + b + c + c + a) ≥ 9,

which follows from Exercise 1
...
For the other inequality use
Exercise 1
...

1
a

+

1
b

≥

4
a+b
...
45
...

n
n

Now, apply the AM-GM inequality
...
46
...

j=i yi ≥ (n − 1)

(n − 1)n
≥

Observe that y1 + · · · +

j=i yj

i
i

yi

j=i yj

1
n−1

and

1
n−1

= (n − 1)n
...
47
...
46 directly
...
48
...
, n + 1
...
Observe that
n

n

n

√
1
ai
1
ai −
√
1 + ai i=1
1 + ai i=1 ai
i=1
√ √
√
√
√
√
( ai aj − 1)( ai − aj )2 ( ai + aj )

...
Hence the
terms of the last sum are positive
...
49
...

n

n

ai −
i=1

Then

bi = 0,
i=1

thus Sa = Sb = S
...
21
...
50
...

Setting x = a3 , y = b3 and z = c3 , the inequality is equivalent to
1
1
1
+
+
≤ 1
...

Now, use that
1
2
x+y+z
xy + yz + zx
≥ (xzy) 3 and
≥ (xyz) 3
...
) is homogeneous if f (ta, tb,
...
) for each t ∈ R
...
) ≥ 0, in the case of a homogeneous function, is equivalent to
f (ta, tb,
...

4
...
Follow the ideas used in the solution of Exercise 1
...
Start with
the inequality (a2 − b2 )(a − b) ≥ 0 to guarantee that a3 + b3 + abc ≥ ab(a + b + c),
then
1
c
≤

...
51
...

1 1 1
1
1
1
1
+ + +
+
+
+
a b
c ab bc ca abc
3
1
3
≥1+ √
+
+
3
abc
3
2
abc
(abc)
=1+

=

3

1
1+ √
3
abc

≥ 43
...
52
...
Now, we use
a
b
c
the AM-GM inequality for each term of the product and the inequality follows
immediately
...
53
...

2
2
2
Solution 1
...
Observe that this exercise is similar to Exercise 1
...

Solution 1
...
Apply the inequality between the arithmetic mean and the harmonic
mean to get
2ab
2
a+b
= 1 1 ≤

...

Solution 1
...
First use the fact that (a + b)2 ≥ 4ab, and then take into account
that
n
n
1
1
≥4
2
...
36 to prove that
n

(ai + bi )
i=1

2

n

1
2

i=1

(ai + bi )

≥ n2
...
57
...
Adding
xy
√
√
similar results we get 2(xy + yz + zx) ≥ 2(x yz + y zx + z xy)
...
Adding similar results
≥
√
√
once more, we obtain x2 + y 2 + z 2 ≥ x yz + y zx + z xy
...

3
+y
Solution 1
...
Using the AM-GM inequality takes us to x4 √ 4 ≥ 2x2 y 2
...
Or, directly we
have that
4 4 4
√
z2 z2
4 x y z
+
≥4
= 8xyz
...
59
...

≥ 2, since (x − 2)2 ≥ 0
...
Let a = x − 1 and b = y − 1, which are positive numbers,
2
2
then the inequality we need to prove is equivalent to (a+1) + (b+1) ≥ 8
...
Then,
2
(a+1)2
b
+ (b+1) ≥ 4 a + a ≥ 8
...
24
...
60
...
2)
...
61
...

1
Observe that ( 1 , 1 , 1 ) and ( a2 , b1 , c1 ) can be ordered in the same way
...
2) to get

1
1
1
+ 3+ 3
a3
b
c
1 1
1 1
1 1
≥ 2 + 2 + 2
a c b a c b
c a
b
= + +
a b
c
= a2 b + b2 c + c2 a
...

Solution 1
...
Use inequality (1
...

c
b

a b c
b , c, a

and

4
...
63
...
2) with (a1 , a2 , a3 ) = (b1 , b2 , b3 ) =
1
(a1 , a2 , a3 ) = 1 , 1 , a
...
64
...
2) twice over with (a1 , a2 , a3 ) = (b, c, a) and
(c, a, b), respectively
...

Solution 1
...
Use the same idea as in the previous exercise, but with n variables
...
66
...

Solution 1
...
Apply Exercise 1
...
, an , a1 ,
...

Solution 1
...
Apply Example 1
...
11
...
69
...

3
3
3
Therefore

1
3

≤ a2 + b2 + c2
...

3

Second solution
...

√
Solution 1
...
Let G = n x1 x2 · · · xn be the geometric mean of the given numbers
x1 x1 x2
2
and (a1 , a2 ,
...
, x1 xG···xn
...
4
...

Also, using Corollary 1
...
2,
n≤

x2
xn
a1
a2
an
x1
+
+ ···+
,
+
+ ··· +
=
an
a1
an−1
G
G
G

then

x1 + x2 + · · · + xn

...

G≤

126

Solutions to Exercises and Problems

Solution 1
...
The inequality is equivalent to
an−1 + an−1 + · · · + an−1 ≥
n
1
2

a 1 · · · an
a1 · · · an
a1 · · · an
+
+ ··· +
,
a1
a2
an

which can be veriﬁed using the rearrangement inequality several times over
...
72
...
Use
i
i
the AM-GM inequality to obtain
1
n

n
i=1

1
√
≥
1 − ai

n
n

≥

i=1

1
√
=
1 − ai
1

1
n

n
i=1 (1

− ai )

1
n
i=1 (1

n

− ai )

n

...

i=1

√
Solution 1
...
(i) 4a + 1 < 4a+1+1 = 2a + 1
...

Solution 1
...
Suppose that a ≥ b ≥ c ≥ d (the other cases are similar)
...
Apply the Tchebyshev inequality twice over to show that
b3
c3
d3
1
1
1
1
a3
1
+
+
+
≥ (a3 + b3 + c3 + d3 )
+ + +
A
B
C
D
4
A B
C
D
1
1
1
1
1 2
(a + b2 + c2 + d2 )(a + b + c + d)
+ + +
≥
16
A B
C
D
=

1 2
(a + b2 + c2 + d2 )
16

1
1
1
1
+ + +
A B
C
D

A+B+C+D
3

...

Solution 1
...
Apply the rearrangement inequality to
⎛
a 3 b 3 c
a
3
,
, (b1 , b2 , b3 ) = ⎝
(a1 , a2 , a3 ) = 3 ,
b
c
a
b

2

,

3

b
c

⎞

2

,

3

c
a

2

⎠

4
...

ab

3

Finally, use the fact that abc = 1
...
The AM-GM inequality and the fact that abc = 1 imply that
a a b
+ +
b
b
c

1
3

≥

3

aab
=
bbc

3

a2
=
bc

3

a3
= a
...

Solution 1
...
Using the hypothesis, for all k, leads to s − 2xk > 0
...

k=1

− 2xk ) = ns − 2s, therefore
n
k=1

x2
s
k

...
77
...

Solution 1
...
The function
f (a, b, c) =

b
c
a
+
+
+ (1 − a)(1 − b)(1 − c)
b+c+1 a+c+1 a+b+1

is convex in each variable, therefore its maximum is attained at the endpoints
...
79
...
By symmetry, the inequality holds for y = 0
...
Let u ≥ 0 and v ≥ 0 such that
x = e−u and y = e−v , then the inequality becomes
1
1
2
√
+√
≤ √
,
−2u
−2v
1+e
1+e
1 + e−(u+v)

128

Solutions to Exercises and Problems

that is,
f (u) + f (v)
≤f
2

u+v
2

,

2x

1
where f (x) = √1+e−2x
...
Thus the previous inequality holds
...
80
...

Solution 1
...
Use log(sin x) or the fact that
sin A sin B = sin

A−B
A+B
+
2
2

sin

A−B
A+B
−
2
2

...
82
...
Suppose
that (1 + nx) > 0
...
, 1, 1 + nx)
with (n − 1) ones
...
, an be positive numbers and deﬁne, for each j = 1,
...

j

Apply Bernoulli’s inequality to show that
which implies
j
j
σj ≥ σj−1 j

σj
− (j − 1)
σj−1

σj
σj−1

j

σ

j
≥ j σj−1 − (j − 1),

j−1
j−1
= σj−1 (jσj − (j − 1)σj−1 ) = aj σj−1
...

Solution 1
...
If x ≥ y ≥ z, we have xn (x − y)(x − z) ≥ y n (x − y)(y − z) and
z n (z − x)(z − y) ≥ 0
...
84
...
The inequality now
follows from Sch¨ r’s inequality
...

Solution 1
...
The inequality is homogeneous, therefore we can assume that a +
x
b + c = 1
...
By Jensen’s inequality
it follows that

a
(1−a)2

+

b
(1−b)2

+

c
(1−c)2

≥ 3f

a+b+c
3

= 3f

1
3

=

3 2
2
...
86
...
Thus, the inequality will hold if

1+

9
6

...
1 Solutions to the exercises in Chapter 1
But this last inequality follows from 1 −
1
a,

129

2
3
a+b+c
= 1 and
b

≥ 0
...
Thus, the inequality is equivalent to
1+

1
c;

it follows immediately

3
6
≥
xy + yz + zx
x+y+z

which is the ﬁrst part of this exercise
...
87
...
First suppose that r > s > 0
...
, xs gives
1
n
r

t1 xr + · · · + tn xr ≥ (t1 xs + · · · + tn xs ) s
1
n
1
n
and taking the 1 -th power of both sides gives the desired inequality
...
Then f (x) = x s is concave, so Jensen’s inequality
is reversed; however, taking 1 -th powers reverses the inequality again
...

c
Solution 1
...
(i) Apply H¨lder’s inequality to the numbers xc ,
...
,
o
n
1
a
b
c
yn with a = c and b = c
...
5
...
The only extra fact that we need to prove is
1
b
c
xi yi zi ≤ a xa + 1 yi + 1 zi , but this follows from part (i) of that example
...
89
...
We have two cases, (i) b ≤ a+b+c and (ii) b ≥ a+b+c
...

3
a+b+c
that
3

It happens
≤ a+c ≤ c, and it is true that
2
there exist λ, μ ∈ [0, 1] such that
c+a
= λc + (1 − λ)
2

a+b+c
3

and

a+b+c
3

≤

b+c
2

b+c
= μc + (1 − μ)
2

≤ c
...

Adding these equalities, we obtain
a + b + 2c
= (λ + μ)c + (2 − λ − μ)
2

a+b+c
3

= (2 − λ − μ)

Hence,
a + b − 2c
= (2 − λ − μ)
2
therefore 2 − (λ + μ) =

3
2

and (λ + μ) = 1
...

130

Solutions to Exercises and Problems
Now, since f is a convex function, we have
a+b
2
b+c
2
c+a
2

f
f
f

≤

1
(f (a) + f (b))
2
a+b+c
3
a+b+c
3

≤ μf (c) + (1 − μ)f
≤ λf (c) + (1 − λ)f

thus, adding these inequalities we get
c+a
2

≤

It is similar to case (i), using the fact that a ≤

a+c
2

f

a+b
2

Case (ii): b ≥

+f

b+c
2

+f

1
(f (a) + f (b) + f (c))
2
3
a+b+c

...

3

≤

a+b+c
3

and a ≤

a+b
2

≤

a+b+c

...
90
...
Applying Popoviciu’s inequality (see the previous exercise) to the function f : R → R+ deﬁned by
f (x) = exp(2x), which is convex since f (x) = 4 exp(2x) > 0, we obtain
exp(2x) + exp(2y) + exp(2z) + 3 exp

2(x + y + z)
3

≥ 2 [exp(x + y) + exp(y + z) + exp(z + x)]
= 2 [exp(x) exp(y) + exp(y) exp(z) + exp(z) exp(x)]
...

For the second part apply the AM-GM inequality in the following way:
√
3
2abc + 1 = abc + abc + 1 ≥ 3 a2 b2 c2
...
91
...

1
1
1
9
4
4
4
We will get the inequality a + b + c + a+b+c ≥ b+c + c+a + a+b
...

Solution 1
...
Observe that by using (1
...
Hence
x2 + y 2 + z 2 − |x||y| − |y||z| − |z||x| =

|xy + yz + zx| ≤ |x||y| + |y||z| + |z||x| ≤ x2 + y 2 + z 2
...
Apply Cauchy-Schwarz inequality to (x, y, z) and (y, z, x)
...
1 Solutions to the exercises in Chapter 1

131

Solution 1
...
The inequality is equivalent to ab + bc + ca ≤ a2 + b2 + c2 , which
we know is true
...
27
...
94
...
7) that
a3 + b3 + c3 = 3abc
...

Solution 1
...
Assume, without loss of generality, that a ≥ b ≥ c
...

Since

−a3 + b3 + c3 + 3abc = (−a)3 + b3 + c3 − 3(−a)bc,

the latter expression factors into
1
(−a + b + c)((a + b)2 + (a + c)2 + (b − c)2 )
...

Solution 1
...
Let p = |(x − y)(y − z)(z − x)|
...
8), we get
3 3 2
p
...

(4
...
2)

Applying again the AM-GM inequality leads to
√
2(x + y + z) ≥ 3 3 p,
and the result follows from inequalities (4
...
2)
...
97
...
7), the condition x3 + y 3 + z 3 − 3xyz = 1 can be
factorized as
(x + y + z)(x2 + y 2 + z 2 − xy − yz − zx) = 1
...
3)
Let A = x2 + y 2 + z 2 and B = x + y + z
...

By identity (1
...
Equation (4
...

B2 − A
2

= 1,

Since B > 0, we may apply the AM-GM inequality to

1
2
1
= B2 + +
≥ 3,
B
B
B
that is, A ≥ 1
...

3A = B 2 +

132

Solutions to Exercises and Problems

Solution 1
...
Inequality (1
...

a
b
c
d
a+b+c+d
a+b+c+d
Solution 1
...
Apply inequality (1
...

a +b =
1
1
2
2
8
4

4

Solution 1
...
Express the left-hand side as
√
√
√
( 2)2
( 2)2
( 2)2
+
+
x+y
y+z
z+x
and use inequality (1
...

Solution 1
...
Express the left-hand side as
y2
z2
x2
+
+
,
axy + bzx ayz + bxy
azx + byz
and then use inequality (1
...
8)
...
102
...
11)
...
103
...
11) to get
x
y
z
(x + y + z)2
+
+
≥ 2

...
1 Solutions to the exercises in Chapter 1

133

but this is equivalent to x2 + y 2 + z 2 ≥ xy + yz + zx
...
11) to get
x
y
z
w
+
+
+
x + 2y + 3z y + 2z + 3w z + 2w + 3x w + 2x + 3y
(w + x + y + z)2

...
This
follows by using the AM-GM inequality six times under the form x2 + y 2 ≥ 2xy
...
104
...
11) to get
y2
z2
x2
+
+
(x + y)(x + z) (y + z)(y + x) (z + x)(z + y)
(x + y + z)2

...

Solution 1
...
We express the left-hand side as
a2
b2
c2
d2
+
+
+
a(b + c) b(c + d) c(d + a) d(a + b)
and apply inequality (1
...

a(b + c) b(c + d) c(d + a) d(a + b)
a(b + 2c + d) + b(c + d) + d(b + c)

134

Solutions to Exercises and Problems

On the other hand, observe that
(a + b + c + d)2
(ac + bd) + (ab + ac + ad + bc + bd + cd)
=

a2 + b2 + c2 + d2 + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd

...

Solution 1
...
We express the left-hand side as
b2
c2
d2
e2
a2
+
+
+
+
ab + ac bc + bd cd + ce de + ad ae + be
and apply inequality (1
...

ab + ac bc + bd cd + ce de + ad ae + be
ab
Since

(a + b + c + d + e)2 =

a2 + 2

ab,

we have to prove that
2

a2 + 4

ab ≥ 5

ab,

which is equivalent to
2
The last inequality follows from

a2 ≥
a2 ≥

ab
...

Solution 1
...
(i) Using Tchebyshev’s inequality with the collections (a ≥ b ≥ c)
2
2
2
and ( a ≥ by ≥ cz ), we obtain
x
1
3

a3
b3
c3
+
+
x
y
z

≥

a2
x

+

b2
y

3

+

c2
z

·

a+b+c
,
3

then by (1
...

x
y
z
x+y+z
Therefore

a3
b3
c3
(a + b + c)2 a + b + c
+
+
≥
·

...
1 Solutions to the exercises in Chapter 1

135

(ii) By Exercise 1
...

Raising to the cubic power both sides and then dividing both sides by 3(x + y + z)
we obtain the result
...
108
...
11), we obtain
x2 + x2 + · · · + x2
1
2
n
x1 + x2 + · · · + xn
x2
x2
x2
1
2
n
=
+
+ ··· +
x1 + x2 + · · · + xn
x1 + x2 + · · · + xn
x1 + x2 + · · · + xn
x1 + x2 + · · · + xn
(x1 + x2 + · · · + xn )2
=

...

Since k = max {x1 , x2 ,
...
, xn } = t, we have that kn ≥ n
t
and since x1 +x2 +···+xn ≥ 1, because all the xi are positive integers, it is enough
n
to prove that
n
x1 + x2 + · · · + xn
≥ x1 · x2 · · · · · xn ,
n
which is equivalent to the AM-GM inequality
...

z
Solution 1
...
Using the substitution a = x , b = y and c = x , the inequality
y
z
takes the form
a3
b3
c3
+ 3
+ 3
≥ 1,
a3 + 2 b + 2 c + 2
and with the extra condition, abc = 1
...

≥ 2
a + b2 + c2 + 2bc + 2ca + 2ab

a3

The inequality above follows from inequality (1
...

136

Solutions to Exercises and Problems

b
Solution 1
...
With the substitution x = a , y = c , z =
b
the form
a
b
c
3
+
+
≥ ,
b+c c+a a+b
2

c
a,

the inequality takes

which is Nesbitt’s inequality (Example 1
...
8)
...
111
...
, xn = an
...

a 1 + a2 + a3
a 2 + a3 + a4
a n + a1 + a2
But this inequality is easy to prove
...
, n
we have
ai + ai+1 + ai+2 < a1 + a2 + · · · + an
...
112
...

x3 + y 3
y + z3
z + x3
Tchebyshev’s inequality can be used to prove that
x3 + y 3 x + y
x4 + y 4
≥
,
2
2
2
thus

x4 + y 4
y4 + z 4
z 4 + x4
x+y y+z z+x
+
+

...
113
...
11)
...

3

Squaring both sides, we obtain 3(xy + yz + zx) ≤ (x + y + z)2 , which is valid if
and only if (xy + yz + zx) ≤ x2 + y 2 + z 2 , something we already know
...
114
...

a+1 b+1 c+1

≤0

4
...

2 + 2yz + z 2 + 2zx + x2
2xy + y
2x
y

The only inequality in the expression follows from inequality (1
...

Solution 1
...
Observe that
[5, 0, 0] =

2 5
2
(a + b5 + c5 ) ≥ (a3 bc + b3 ca + c3 ab) = [3, 1, 1],
6
6

where Muirhead’s theorem has been used
...
116
...

2
2
2
2

This is equivalent to
(a2 + b2 + c2 )2 ≥ 3[((a + b)2 − c2 )(c2 − (b − a)2 )]
= 3(2c2 a2 + 2c2 b2 + 2a2 b2 − (a4 + b4 + c4 )),
that is, a4 + b4 + c4 ≥ a2 b2 + b2 c2 + c2 a2 , which, in terms of Muirhead’s theorem,
is equivalent to proving [4, 0, 0] ≥ [2, 2, 0]
...
Using the substitution
x = a + b − c, y = a − b + c, z = −a + b + c,
we obtain x + y + z = a + b + c; then, using Heron’s formula we get
4(ABC) =

(a + b + c)(xyz) ≤

(a + b + c)

(a + b + c)2
(x + y + z)3
√
=

...
This last inequality
follows from Muirhead’s theorem, since [1, 1, 0] ≤ [2, 0, 0]
...
117
...

Solution 1
...
The inequality is equivalent to
a3 + b3 + c3 ≥ ab(a + b − c) + bc(b + c − a) + ca(c + a − b)
...
Then, the inequality we have to prove is
2

z+x
2 ,

b=

x+y
2 ,

1
1
((z+x)3 +(x+y)3 +(y+z)3) ≥ ((z+x)(x+y)x+(x+y)(y+z)y+(y+z)(z+x)z),
8
4
which is again equivalent to
3(x2 y + y 2 x + · · · + z 2 x) ≥ 2(x2 y + · · · ) + 6xyz
or

x2 y + y 2 x + y 2 z + z 2 y + z 2 x + x2 z ≥ 6xyz,

and applying Muirhead’s theorem we obtain the result when x, y, z are nonnegative
...

Solution 1
...
Observe that
a3
b3
c3
+ 2
+ 2
≥a+b+c
b2 − bc + c2
c − ca + a2
a − ab + b2
is equivalent to the inequality
a3 (b + c) b3 (c + a) c3 (a + b)
+ 3
+ 3
≥ a + b + c,
b 3 + c3
c + a3
a + b3
which in turn is equivalent to
a3 (b + c)(a3 + c3 )(a3 + b3 ) + b3 (c + a)(b3 + c3 )(a3 + b3 )
+ c3 (a + b)(a3 + c3 )(b3 + c3 )
≥ (a + b + c)(a3 + b3 )(b3 + c3 )(c3 + a3 )
...
1 Solutions to the exercises in Chapter 1

139

The last inequality can be written in the terminology of Muirhead’s theorem as
[9, 1, 0] + [6, 4, 0] + [6, 3, 1] + [4, 3, 3] ≥

1
[1, 0, 0]
2

1
[6, 3, 0] + [3, 3, 3]
3

= [7, 3, 0] + [6, 4, 0] + [6, 3, 1] + [4, 3, 3]
⇔ [9, 1, 0] ≥ [7, 3, 0],
a direct result of Muirhead’s theorem
...
120
...

(1 + b) (1 + c)
(1 + c) (1 + a)
(1 + a) (1 + b)
Use Tchebyshev’s inequality to prove that
a3
b3
c3
+
+
(1 + b) (1 + c) (1 + c) (1 + a) (1 + a) (1 + b)
1
1
1
1
+
+
≥ (a3 + b3 + c3 )
3
(1 + b)(1 + c) (1 + a)(1 + c) (1 + a)(1 + b)
3 + (a + b + c)
1
= (a3 + b3 + c3 )

...

8

≥ 1
...
Multiplying by the common denominator and expanding both
sides, the desired inequality becomes
4(a4 + b4 + c4 + a3 + b3 + c3 ) ≥ 3(1 + a + b + c + ab + bc + ca + abc)
...

Now, note that
[4, 0, 0] ≥

4 4 4
4 4 4
, ,
= a 3 b 3 c 3 = 1 = [0, 0, 0],
3 3 3

140

Solutions to Exercises and Problems

where it has been used that abc = 1
...

3

Finally, [3, 0, 0] ≥ [1, 1, 1]
...

4
...
1
...

(iii) a = x + y, b = y + z, c = z + x ⇔ x = a+c−b , y = a+b−c , z = b+c−a
...
2
...

(ii) With 2, 3 and 4 it is possible to construct a triangle but with 4, 9 and 16 it is
not possible to do so
...

c
Solution 2
...
Use the fact that if a, b, c are the lengths of the sides of a triangle,
the angle that is opposed to the side c is either 90◦ or acute or obtuse if c2 is equal,
less or greater than a2 + b2 , respectively
...

Solution 2
...
Since ∠A > ∠B then BC > CA
...

Solution 2
...
(i) Let O be the intersection point of the diagonals AC and BD
...
Adding the inequalities, we get AB + CD < AC + BD
...
Adding these last two inequalities we get AB < AC
...
Refer to the previous exercise
...
2 Solutions to the exercises in Chapter 2

141

Solution 2
...
Each di is less than the sum of the lengths of two sides
...

Solution 2
...
Use the triangle inequality in the triangles ABA and AA C to prove
that c < ma + 1 a and b < ma + 1 a
...
8
...
Therefore,
180◦ = α + β + γ > α1 + α2 + α = 2α
...

Solution 2
...
Construct a parallelogram ABDC, with one diagonal BC and the
other AD which is equal to two times the length of AA and use D2 on the triangle
ABD
...
10
...
Similarly, mb < a+c and mc < a+b
...

A

B

C

B

A

A

C

Extend the segment C B to a point A such that C A = BC
...

Solution 2
...
Consider the quadrilateral ABCD and let O be a point on the
exterior of the quadrilateral so that AOB is similar to ACD, and thus OAC and
BAD are also similar
...
17
Solution 2
...
Set a = AB, b = BC, c = CD, d = DA, m = AC and n = BD
...
Thus we have18
m(ab + cd)
,
4R
n(bc + ad)
(ABCD) = (BCD) + (DAB) =

...

[6, page 97] or [9, page 13]
...

Solution 2
...
Apply to the triangle ABP a rotation of 60◦ with center at A
...
The triangle P P C has as sides P P = P A, P C = P B and P C, and then the
result
...
Apply Ptolemy’s inequality (see Exercise 2
...

Third solution
...
Let P be the point where AP
intersects the side BC
...
In a
similar way, the other inequalities P B < P C + P A and P C < P A + P B hold
...
14
...
Remember that in
a paralelogram we have 2(a2 + b2 ) = x2 + y 2
...
It is clear that 2b < x + y, therefore (2b)2 < (x + y)2 =
x2 + y 2 + 2xy = 2(a2 + b2 ) + 2xy
...

Solution 2
...
(i) Extend the medians AA , BB and CC until they intersect the
circumcircle at A1 , B1 and C1 , respectively
...
Also, use the facts that ma + A A1 ≤ 2R and that the length of
2

2

2

the median satisﬁes m2 = 2(b +c )−a , that is, 4m2 + a2 = 2(b2 + c2 )
...

(ii) Use Ptolemy’s inequality in the quadrilaterals AC GB , BA GC and CB GA ,
where G denotes the centroid
...

4
...
16
...
Now, using the triangle inequality, prove that mb + mc < 2 (b + c)
...

The right-hand side inequality can be obtained from the ﬁrst when applied
to the triangle of sides19 with lengths ma , mb and mc
...
17
...
If E and F are the
projections of Ia on the sides AB and CA, respectively, it is clear that if ra is the
radius of the excircle, we have that ra = Ia E = EA = AF = F Ia = s, where s is
the semiperimeter of ABC
...
Since aha = bc, we have that
r
ha
bc
AD
=
=
=
DIa
ra
as

abc
4R

4Rr
a2

1
rs

=

4Rr
,
a2

where r and R are the inradius and the circumradius of ABC, respectively
...
Then, it is enough to
a
a
√
√
2
2
2
prove that b+c ≤ 2 or, equivalently, that 2bc ≤ a2 , but bc = b2 c2 ≤ b +c = a
...
18
...
27
...

Solution 2
...
Use the previous suggestion
...
20
...

Solution 2
...
The ﬁrst inequality is the Nesbitt’s inequality, Example 1
...
8
...

2
Solution 2
...
Observe that a2 (b + c − a) + b2 (c + a − b) + c2 (a + b − c) − 2abc =
(b + c − a) (c + a − b) (a + b − c), now see Example 2
...
3
...
23
...
22
...
24
...

Then, the inequality is equivalent to
(1
...

19 See

the solution of Exercise 2
...

x2
y

2

2

+ y + z ≥ x+y +z
...
25
...

(a + b)(b + c)(c + a)
8
For the last inequality, see the solution of Example 2
...
3
...
26
...
18,
3(ab + bc + ca) ≤ (a + b + c)2 ≤ 4(ab + bc + ca)
...

Solution 2
...
Use Ravi’s transformation, a = y + z, b = z + x and c = x + y
...

=
2
xyz
2r
For the last identity, see the end of the proof of Example 2
...
4
...
28
...

For (ii), use Ravi’s transformation, a = y + z, b = z + x, c = x + y, in order
to see that the inequality is equivalent to
4(xy + yz + zx) ≤ (y + z)(z + x) + (z + x)(x + y) + (x + y)(y + z)
...
27
...
2 Solutions to the exercises in Chapter 2

145

Another way to obtain (ii) is the following: the given inequality is equivalent
to 3s2 − 2s(a + b + c) + (ab + bc + ca) ≤ ab+bc+ca , which in turn is equivalent to
4
3(ab + bc + ca) ≤ 4s2
...

Solution 2
...
Applying the cosine law, we can see that
a 2 + b 2 − c2

a 2 − b 2 + c2 =

√
√
2ab cos C 2ac cos B

= 2a

(b cos C)(c cos B)
b cos C + c cos B
= a2
...
30
...

Therefore
a 2 + b 2 − c2

a 2 − b 2 + c2 =

cyclic

1
2

a 2 + b 2 − c2
cyclic

c2 + a2 − b 2

+
≤

1
2

a 2 − b 2 + c2

c2 − a2 + b 2

(2a2 )(2c2 ) =
cyclic

ac
...
31
...
The inequalities are equivalent to proving that
z+x x+y
y+z
+
+
≥3
2x
2y
2z

2x
2y
2z
+
+
≥ 3
...

y
x

+

x
y

≥ 2 and for the second inequality

Solution 2
...
Since in triangles with the same base, the ratio between its altitudes
is equal to the ratio of theirs areas, we have that
PQ
PR
PS
(P BC) (P CA) (P AB)
(ABC)
+
+
=
+
+
=
= 1
...
3) of Section 2
...

1
1
1
Solution 2
...
(i) Recall that (S1 + S2 + S3 )( S1 + S2 + S3 ) ≥ 9
...

146

Solutions to Exercises and Problems

Using the formula for the area that is related to the sine of the angle, prove that
S1 S2 S3 = T1 T2 T3
...

3
S1 S2 S3

1
1
1
+
+
S1
S2
S3

The equality holds when the point O is the centroid of the triangle and the lines
through O are the medians of the triangle; in this case S1 = S2 = S3 = T1 = T2 =
T3 = 1 S
...
34
...

BP
CP
On the other hand, if AP + P M + P N = 6, we have AL + BM + CN = 9
...
This implies that

CN
AL BM
+
+
PL PM
PN

PL PM
PN
+
+
AL BM
CN

= 9
...
3), the equality above holds only in the case when
CN
P N = 3, which implies that P is the centroid
...
35
...
20 Thus, the solution follows from part (i) of Example 2
...
4
...
3
...

AD
BE
CF
AD
BE
CF
Since

AD
AD

+

BE
BE

+

CF
CF

AD
AD

+

BE
BE

+

CF
CF

≥ 9, we have the result
...
36
...
3
...

(b + c)2

2
Since 4bc ≤ (b + c)2 , it follows that la ≤ s(s − a) and la lb ≤ s (s − a)(s − b) ≤
(s−a)+(s−b)
c
s
= s2
...

2
20 See

[6, page 85] or [9, page 37]
...
2 Solutions to the exercises in Chapter 2

147

Solution 2
...
Let α = ∠AM B, β = ∠BN A, γ = ∠AP C, and let (ABC) be the
area of ABC
...
Similarly,

abc
1
a · AM sin α =

...
Thus,

bc
ca
ab
+
+
= 2R(sin α + sin β + sin γ) ≤ 6R
...

Solution 2
...
Let A1 , B1 , C1 be the midpoints of the sides BC, CA, AB, respectively, and let B2 , C2 be the reﬂections of A1 with respect to AB and CA,
respectively
...
Then,
2DE = B2 C2 ≤ C2 B1 + B1 C1 + C1 B2 = A1 B1 + B1 C1 + C1 A1 = s
...
Then,
a
s ≥ 2DE = 2ma sin A = 2ma 2R = ama , that is, ama ≤ sR
...

Solution 2
...
The inequality is equivalent to 8(s − a)(s − b)(s − c) ≤ abc, where
s is the semiperimeter
...

Solution 2
...
The area of a triangle ABC satisﬁes the equalities (ABC) = abc =
4R
(a+b+c)r
1
1
1
1
1
, therefore ab + bc + ca = 2Rr ≥ R2 , where R and r denote the circumradius
2
and the inradius, respectively
...
41
...
40 and the sine law
...
42
...

21 Notice

that sin2

1−
1 − cos A
A
=
=
2
2

b2 +c2 −a2
2bc

2

=

a2 − (b − c)2
(s − b)(s − c)
=

...
43
...
3), we know that
1 1 1
+ +
a b
c

(a + b + c)

≥ 9
...

a b
c
R

(4
...
3), we get
π
π
1 π
+
+
3 2A 2B
2C
π
Let f (x) = log 2x , since f (x) =
we get

1
x2

≥

2
π (A

3
3
=
...
5)

> 0, f is convex
...

Applying (4
...

2

We can suppose that a ≤ b ≤ c, which implies A ≤ B ≤ C
...
Using Tchebyshev’s inequality,
π
1
π
1
π
1
log
+ log
+ log
≥
a
2A b
2B
c
2C

1 1 1
+ +
a b
c

(4
...

Therefore, using (4
...
6) leads us to
√
1
3
π
1
π
1
π
3
log
+ log
+ log
≥
log
...
In all the above inequalities, the equality holds
if and only if a = b = c (this means, equality is obtained if and only if the triangle
is equilateral)
...
44
...
2 Solutions to the exercises in Chapter 2

149

where a, b, c are the lengths of the sides of the triangle and R is the circumradius
of the triangle
...

Solution 2
...
Use Leibniz’s inequality and the fact that the area of a triangle is
given by (ABC) = abc
...
46
...

Applying Leibniz’s inequality to DEF , we get
EF 2 + F D2 + DE 2 ≤ 9r2 ,
where r is the inradius of ABC
...
4
...

3
Solution 2
...

a2
b2
c2
a2 bc + b2 ca + c2 ab
abc(a + b + c)
+
+
=
=
hb hc
hc ha
ha hb
4(ABC)2
4(ABC)2
abc(a + b + c)
2R
≥ 4
...
48
...
5
...

2

A
2

=

1−cos A
2

and use that cos A + cos B +

Solution 2
...
Observe that
√
4 3(ABC) ≤

√
√
9abc
9 · 4 Rrs
2s
⇔ 4 3rs ≤
⇔ 2 3s ≤ 9 R ⇔ √ ≤ R
...
4
...

Solution 2
...
Use the previous exercise and the inequality between the harmonic
mean and the geometric mean,
3
1
ab

+

1
bc

+

1
ca

≤

√
3
a 2 b 2 c2
...
51
...

a 2 b 2 c2 ≤
3
Solution 2
...
First, observe that if s =

a+b+c
,
2

then

a2 + b2 + c2 −(a − b)2 − (b − c)2 − (c − a)2 =
= a2 − (b − c)2 + b2 − (c − a)2 + c2 − (a − b)2
= 4{(s − b)(s − c) + (s − c)(s − a) + (s − a)(s − b)}
...

Squaring and simplifying the last inequality, we get
xyz(x + y + z) ≤ x2 y 2 + y 2 z 2 + z 2 x2
...

Solution 2
...
Use Exercise 2
...

Solution 2
...
Note that
9abc
3(a + b + c)abc
≥
ab + bc + ca
a+b+c

⇔ (a + b + c)2 ≥ 3(ab + bc + ca)
⇔

a2 + b2 + c2 ≥ ab + bc + ca,

now, use Exercise 2
...

Solution 2
...
Using (2
...
6) and (2
...

Solution 2
...
Observe the relationships used in the proof of Exercise 2
...

4Rs(rs)
R

4
...
57
...

2
Solution 2
...
Start on the side of the equations which expresses the relationship
between the τ ’s and perform the operations
...
59
...
are the lengths into which each side
is divided for the corresponding point, we can deduce that a2 + b2 + c2 + d2 =
(x2 + (1 − xi )2 )
...

i
i
2
2
2
For part (ii), the inequality on the right-hand side follows from the triangle
inequality
...

a

d
b

c

Solution 2
...
This is similar to part (ii) of the previous problem
...
61
...
Let X, Y , Z be the tangency points of the incircle
with the sides BC, CA, AB, respectively, and let p = a + b + c be the perimeter of
the triangle ABC
...

a
p
p
Similarly, we have the other relations
FG
2z
=
,
c
p

HI
2y
=

...
Thus,
3
p(DEF GHI) ≤ 2(a + b + c) − 4 (a + b + c) = 2 (a + b + c)
...
62
...

The circles with centers the midpoints of the sides of √ triangle and radii 1 cover
the
a circle of radius 2
...

Solution 2
...
Take the acute triangle with sides of lengths 2r1 , 2r2 and 2r3 , if it
exists
...
If the triangle does not exist, the maximum
radius between r1 , r2 and r3 is the answer
...
64
...

Lemma 1
...

d
a
c

4
...
Since
the parallel lines form a square inside the rectangle and such a square contains
the original square, we have the result
...
The diagonal of a square inscribed in a right triangle is less than or
equal to the length of the internal bisector of the right angle
...

It can be assumed that the vertices P and Q belong to the legs of the right
triangle (otherwise, translate the square) and let O be the intersection point of
the diagonals P R and QS
...
Let T be the intersection of
BO with RS, then ∠QBT = ∠QST = 45◦ , therefore BQT S is cyclic and the
center O of the circumcircle of BQT S is the intersection of the perpendicular
bisectors of SQ and BT
...
Since O O > O V , then the chords SQ and BT satisfy SQ < BT ,
and the lemma follows
...
Let ABCD be the square of side
1 and let l be a line that separates the two squares
...
Otherwise, l intersects every
line that determines a side of the square ABCD
...

154

Solutions to Exercises and Problems

D

A

F
E

a

b

B
G

H

C

If l intersects the sides of ABCD in E, F , G, H as in the ﬁgure, we have, by
Lemma 2, that the sum of the√
lengths of the diagonals of the small squares is less
√
than or equal to AC, that is, 2(a + b) ≤ 2, then the result follows
...
65
...
Therefore,
2
2
2
β
γ
α+β+γ
α
3
3
◦
= 8 sin (30 ) = 1,
abc = 8 sin sin sin ≤ 8 sin
2
2
2
6
where the inequality follows from Exercise 1
...

Solution 2
...
The ﬁrst observation that we should make is to check that the
diagonals are parallel to the sides
...
Now, the pentagon can be divided into
(ABCDE) = (ABC) + (ACX) + (CDE) + (EAX)
...
Let a =
(CXA)
AX
(CDX) = (EAX) and b = (DEX), then we get a = XD = (CDX) = a+b , that
b
a
is,

a
b

=

√
1+ 5
2
...

Solution 2
...
Prove that sr = s1 R = (ABC), where s1 is the semiperimeter of
the triangle DEF
...
Use also
that R ≥ 2r
...
68
...
Now, use
√
the fact that (ABC) = 2hb hc and that 23 ≤ sin A ≤ 1
...

4
...
69
...

(i) (ABCD) = (ABC) + (CDA) = ab sin B + cd sin D ≤ ab+cd
...
The quadrilaterals ABCD and ABC D
have the same area but the second one has sides of length a, c, b and d, in this
order
...

(iii) (ABC) ≤ ab , (BCD) ≤ bc , (CDA) ≤ cd and (DAB) ≤ da
...
70
...
7
...

2r

Use the AM-GM inequality
...
71
...

≥ 3
...

For the last inequalities in (i) and (iii), we have used Exercise 2
...
For the
last inequality in (ii), we have resorted to Example 2
...
6
...
7
...
Let A , B , C be the inverses of A,
B, C, respectively
...

Let us prove that pa = pa P B2·P C
...
Similarly, pb = pb P C 2·P A
d
and pc = pc P A2·P B
...

Now, since P A · P A = P B · P B = P C · P C = d2 , after substitution we get
1
1
pb
pc
1
pa
+
+
≥2
+
+
PA PB
PC
PB · PC
PC · PA PC · PA
and this inequality is equivalent to
P B · P C + P C · P A + P A · P B ≥ 2(pa P A + pb P B + pc P C)
...
7
...

156

Solutions to Exercises and Problems

Solution 2
...
If P is an interior point or a point on the perimeter of the triangle
ABC, see the proof of Theorem 2
...
2
...

Since ha ≤ P A + pa (even if pa ≤ 0, that is, if P is a point on the outside of
the triangle, on a diﬀerent side of BC than A), and because the equality holds if P
is exactly on the segment of the altitude from the vertex A, therefore aP A+ apa ≥
aha = apa + bpb + cpc , hence aP A ≥ bpb + cpc
...

Similarly, bP B ≥ apc + cpa and cP C ≥ apb + bpa , therefore
PA + PB + PC ≥

b c
+
c b

pa +

a
c
+
pb +
a
c

b
a
+
b
a

pc
...

Second solution
...
Let H and G be the orthogonal
projections of B and C, respectively, over the segment M N
...

Since ∠BN H = ∠AN M = ∠AP M , the right triangles BN H and AP M are
similar, therefore HN = P M BN
...

PA
PA
Applying Ptolemy’s theorem to AM P N , we obtain P A · M N = AN · P M +
AM · P N , hence
MN =

AN · P M + AM · P N
,
PA

from there we get
BC ≥

AN · P M + AM · P N
PN
PM
BN +
+
CM
...

c
b
Then, P A ≥ pb a + pc a
...

Solution 2
...
Take a sequence of reﬂections of the quadrilateral ABCD, as shown
in the ﬁgure
...
2 Solutions to the exercises in Chapter 2

157
P

A

B

B
S

D

C
R

S

R

R

D
C

Q

A
A

P

B

Note that the perimeter of P QRS is the sum of the lengths of the piecewise line
P QR S P
...

Solution 2
...
First note that (DEF ) = (ABC) − (AF E) − (F BD) − (EDC)
...

(ABC)
cb
(ABC)
ac
(ABC)
ba
Therefore,
(DEF )
z
y
x
z
y
x
=1−
1−
−
1−
−
1−
(ABC)
c
b
a
c
b
a
x
y
z
x y z
x y z
= 1−
1−
1−
+ · · =2 · ·
...
Now, the last product is maximum when x = y =
a
b
z
1
c , and since the segments concur the common value is 2
...

Solution 2
...
If x = P D, y = P E and z = P F , we can deduce that 2(ABC) =
ax + by + cz
...

(a+b+c)2
2(ABC)

b
c
a
+ +
x y z

(ax + by + cz)
...
76
...

Now, (BD + DC)2 = a2 , hence BD2 + DC 2 = a2 − 2BD · DC
...
Thus, BD2 + DC 2 + CE 2 + AE 2 + AF 2 + F B 2 = a2 + b2 +
c2 − 2(BD · DC + CE · AE + AF · F B)
...
But BD·DC ≤ BD+DC = a and the maximum is attained when
2
2
BD = DC
...

Solution 2
...
Since

3

(aP D)(bP E)(cP F ) ≤
8 (ABC)
27
abc

3

aP D+bP E+cP F
3

=

2(ABC)
,
3

we can

...

But c · P F = b · P E ⇔ (ABP ) = (CAP ) ⇔ P is on the median AA
...

Solution 2
...
Using the technique for proving Leibniz’s theorem, verify that
3P G2 = P A2 + P B 2 + P C 2 − 1 (a2 + b2 + c2 ), where G is the centroid of ABC
...

Solution 2
...
The quadrilateral AP M N is cyclic and it is inscribed in the circle of
diameter AP
...
The biggest circle will be attained when the diameter AP is the
biggest possible
...
In this case M

A
N
P

M
B

C

and N coincide with B and C, respectively
...

Solution 2
...
The circumcircle of DEF is the nine-point circle of ABC, therefore
it intersects also the midpoints of the sides of ABC and goes through L, M , N ,
the midpoints of AH, BH, CH, respectively
...
2 Solutions to the exercises in Chapter 2

159

A
L E
F
H

N

M
O
∠A

B

D

A

C

Note that t2 = AL · AD, then
a
t2
a
=
ha

AL · AD
=
AD

AL =

R cos A ≤ 3R cos

=

OA

A+B+C
3
= 3R cos 60◦ = R
...
See Lemma 2
...
2
...
81
...

=
2(ABC)
Now use the fact that
pa
pb
pc
+
+
ha
hb
hc

ha
hb
hc
+
+
pa
pb
pc

≥ 9
...

(iii) Let x = (P BC), y = (P CA) and z = (P AB)
...
Similarly, we have that b(hb − pb ) ≥ 4 zx y
√
c(hc − pc ) ≥ 4 xy
...

Therefore, (ha − pa )(hb − pb )(hc − pc ) ≥ 8pa pb pc
...
82
...
If E is the projection of I on AD, it is enough to prove that AE ≥ AO = R
...

If I is projected on E in the diameter AA , then AE = AE
...

To see that COF is an acute triangle, use that the angles of ABC satisfy
∠A < ∠B < ∠C, so that 1 ∠B < 90◦ − ∠A, 1 ∠C < 90◦ − ∠A
...

Solution 2
...
Let ABC be a triangle with sides of lengths a, b and c
...

2

(4
...
Then the product 16(ABC)2 is
maximum when (s − a)(s − b) is maximum, that is, if s − a = s − b, which is
equivalent to a = b
...

Solution 2
...
Let ABC be a triangle with sides of length a, b and c
...
Using (4
...
The product of
these three numbers is maximum when (s − a) = (s − b) = (s − c), that is, when
a = b = c
...

Solution 2
...
If a, b, c are the lengths of the sides of the triangle, observe that
a+b+c = 2R(sin ∠A+sin ∠B +sin ∠C) ≤ 6R sin ∠A+∠B+∠C , since the function
3
sin x is concave
...

Solution 2
...
The inequality (lm + mn + nl)(l + m + n) ≥ a2 l + b2 m + c2 n is
equivalent to
l 2 + m 2 − c2
m2 + n 2 − b 2
n 2 + l 2 − a2
+
+
+3≥0
lm
mn
nl
3
⇔ cos ∠AP B + cos ∠BP C + cos ∠CP A + ≥ 0
...

2
2

3
2

≥ 0 is equivalent to (2 cos α+β +
2

Solution 2
...
Consider the Fermat point F and let p1 = F A, p2 = F B and
p3 = F C, then observe ﬁrst that (ABC) = 1 (p1 p2 + p2 p3 + p3 p1 ) sin 120◦ =
2
√
3
(p1 p2 + p2 p3 + p3 p1 )
...

1
2
3

4
...
Then, a2 + b2 + c2 ≥ 4 3(ABC)
...

Solution 2
...
Let a, b, c be the lengths of the sides of the triangle ABC
...
From the solution of the previous exercise we know that
√
4 3(ABC) = 3(p1 p2 + p2 p3 + p3 p1 )
...
27
...
89
...

In the ﬁrst case the minimum of P A + P B + P C is CC , where C is the
image of A when we rotate the ﬁgure in a positive direction through an angle of
60◦ having B as the center
...

2
√
√
Now, use the fact that a2 + b2 + c2 ≥ 4 3(ABC) to obtain (CC )2 ≥ 4 3(ABC)
...
4
...

When ∠A ≥ 120◦ , the point that solves Fermat-Steiner problem is the point
A, then P A + P B + P C ≥ AB + AC = b + c
...

xyz
Moreover, we can use the fact that b = x + z, c = x + y and r = x+y+z
...
It is clear that P A + pa ≥ ha , where pa is the distance from
P to BC and ha is the length of the altitude from A
...

o
1
1
1
Now using Exercise 1
...
Therefore, 9r ≤ ha + hb + hc ≤ 2 (P A + P B + P C) and the
result follows
...
90
...
Since
2
P B1 CA1 is a cyclic quadrilateral with diameter P C, applying the sine law leads
us to A1 B1 = P C sin C
...

162

Solutions to Exercises and Problems

Call Q the intersection of BP with the circumcircle of triangle ABC, then
∠B1 A1 C1 = ∠QCP
...
Similarly, ∠C1 BP = ∠C1 A1 P
...
Therefore,
∠B1 A1 C1 = ∠B1 CP + ∠ACQ = ∠QCP
...

PC
1
A1 B1 · A1 C1 sin ∠B1 A1 C1
2
1
= P B · P Csin B sin C sin ∠QCP
2
1
PQ
= P B · P C · sin B sin C
sin ∠BQC
2
PC

(A1 B1 C1 ) =

=

1
P B · P Q · sin A sin B sin C
2

=

(R2 − OP 2 )(ABC)

...
The area of A1 B1 C1 is maximum when P = O, that is,
when A1 B1 C1 is the medial triangle
...
3 Solutions to the problems in Chapter 3
Solution 3
...
Let a = A1 A2 , b = A1 A3 and c = A1 A4
...

4
...
Similarly
(b+c)2
1
2(b+c)2 = 2
...

B1 B2
B1 B3

A1 A2
a
A1 A3 = b
+S
Therefore SBSA C =

=

+
=
=
>
The third equality follows from
ab + ac = bc and the inequality follows from inequality (1
...
The inequality is
strict since b = c
...

b
c
b
c
bc
bc
The sine law applied to the triangle A1 A3 A4 leads us to
sin2 π
sin2 π
a2
7
7
=
2π
4π =
2π
bc
sin 7 sin 7
2 sin 7 sin 2π cos 2π
7
7
=

a2
b2

sin2 π
7
4 cos 2π (1 + cos 2π ) sin2
7
7

1
4 cos
+ cos 2π )
7
√
2−1
1
1
√
= √

...

c
bc
=

Thus

sin2 π
sin2 π
7
7
=
2(1 − cos2 2π ) cos 2π
2 cos 2π (1 + cos 2π )(1 − cos 2π )
7
7
7
7
7
π
7

=

2π
7 (1

Solution 3
...
Cut the tetrahedron along the edges AD, BD, CD and place it on
the plane of the triangle ABC
...
Observe that D3 , A and D1 are
collinear, as are D1 , B and D2
...

Then AB = 1 D2 D3 and by the triangle inequality, D2 D3 ≤ CD3 + CD2 = 2CD
...

Solution 3
...
Letting S be the area of the triangle, we have the formulae sin α = 2S ,
bc
2S
sin β = 2S , sin γ = 2S and r = S = a+b+c
...

Solution 3
...
Suppose that the circles have radii 1
...

We have to minimize the common area between any pair of circles, which will be
minimum if the point P is in the interior of the triangle ABC (otherwise, rotate
one circle by 180◦ around P , and this will reduce the common area)
...
It is clear that

164

Solutions to Exercises and Problems

A
α
β
B

P
βγ

α
γ
C

α+β +γ = 180◦
...

Solution 3
...
Let I be the incenter of ABC, and draw the line through I perpendicular to IC
...

First prove that (CDE) ≥ (CD E ) by showing that the area of D DI is greater
than or equal to the area of EE I; to see this, observe that one of the triangles
DD I, EE I lies in the opposite side to C with respect to the line D E , if for
instance, it is DD I, then this triangle will have a greater area than the area of
EE I, then the claim
...

=
C
C
2
sin C
2 sin 2 cos 2
√
Solution 3
...
The key is to note that 2AX ≥ 3(AB +BX), which can be deduced
by applying Ptolemy’s inequality (Exercise 2
...
Hence
(CD E ) =

√
2AD = 2(AX + XD) ≥ 3(AB + BX) + 2XD
√
√
≥ 3(AB + BC + CX) + 3XD
√
≥ 3(AB + BC + CD)
...
7
...
Construct another triangle ABC that has A B C as its

4
...
In ABC we can ﬁnd
all the points
...
For instance, if Q is in the half-plane determined by
BC, opposite to where A lies, the triangle QB C has greater area than A B C ,
a contradiction
...
8
...
Let us prove that M is the desired minimum
2
value, which is achieved by setting x1 = x2 = · · · = xn = 1
...
Therefore
k

x1 +

xn
x2 x3
1
n−1
2
+ 3 +· · ·+ n ≥ x1 +x2 − +· · ·+xn −
= x1 +x2 +· · ·+xn −n+M
...

We conclude that the given expression is at least n − n + M = M
...

Second solution
...

j

= n
...
9
...
Let C and F
be points outside the hexagon and such that ABC and DEF are also equilateral
triangles
...
Now use the fact that AC BG and
EF DH are cyclic in order to conclude that AG + GB = GC and DH + HE =
HF
...
10
...
Since
9
abc
rs = abc , we can deduce that 2rR = a+b+c
...

Solution 3
...
The left-hand side of the inequality follows from
√
1
1 + x0 + x1 + · · · + xi−1 xi + · · · + xn ≤ (1 + x0 + · · · + xn ) = 1
...
, n
...

sin θi −sin θi−1
cos θi−1

It is left to prove that

sin θi − sin θi−1 = 2 cos

<

π
2
...

2
2

To show the inequality, use the facts that cos θ is a decreasing function and that
sin θ ≤ θ for 0 ≤ θ ≤ π
...

2

2
Solution 3
...
If i=1 xi = 1, then 1 = ( i=1 xi ) =
i=1 xi + 2
Therefore the inequality that we need to prove is equivalent to
n

n

1
≤
n−1

n
i=1

n

i
xi xj
...

1 − ai

Use the Cauchy-Schwarz inequality to prove that
2

n

xi

n

≤

i=1

i=1

x2
i
1 − ai

n

(1 − ai )
...
13
...
The inequality
n+1
that we need to prove is reduced to
n

n

xi (xn+1 − xi ) ≤

√

n − 1xn+1
...
, xn ) and ( xn+1 − x1 ,
...

Solution 3
...
First, recall that N is also the midpoint of the segment that joins
the midpoints X and Y of the diagonals AC and BD
...

Solution 3
...
The inequality on the right-hand side follows from wx + xy + yz +
zw = (w + y)(x + z) = −(w + y)2 ≤ 0
...
3 Solutions to the problems in Chapter 3

167

For the inequality on the left-hand side, note that
|wx + xy + yz + zw| = |(w + y)(x + z)|
1
2
2
(w + y) + (x + z)
≤
2
≤ w2 + x2 + y 2 + z 2 = 1
...

Solution 3
...
For the inequality on the left-hand side, rearrange as follows:
a n + a2
a 1 + a3
an−1 + a1
a1
a2
a3
a2
a1
an
+
+ ···+
=
+
+
+
+ ···+
+
,
a1
a2
an
a2
a1
a2
a3
an
a1
now, use the fact that

y
x
y + x ≥ 2
...
Using induction, prove that
a
Sn ≤ 3n
...
If a = b = c,
a
b
c
then b+c + c+a + a+b = 6 and the inequality is true
...
In all of them, we have a ≤ b and a < c
...

b
b
Thus, b+c + c+a + a+b = a+2b + 2a+b + 1 = 3 + 2 a + 2 a
...
This means the sum is at
most 8, which is less than 9, then the result
...
Suppose that Sn−1 ≤ 3(n − 1)
...
, an }, if all are equal, then Sn = 2n and the inequality is true
...
Take the maximum of the ai ’s;
its neighbors (ai−1 , ai+1 ) can be equal to this maximum value, but since there are
two diﬀerent numbers between the ai ’s for some maximum ai , we have that one of
its neighbors is less than ai
...
Then,
+a
+a
since 2an > an−1 + a1 , we have that an−1n 1 < 2 and then an−1n 1 = 1, for which
a
a
an = an−1 + a1
...

a1
a2
an−1

Sn =

Since Sn−1 ≤ 3(n − 1), this implies that Sn ≤ 3n
...
17
...
On the
BC
BC
other hand, since the triangles BCE and DCA are similar, as well as the triangles
ABD and F BC, it happens that R BD + DC = R AD + AD
...

AD
CF
Multiplying these equalities and applying the AM-GM inequality, the result is
attained
...
Let D , E and F be the
intersection points of AO, BO and CO with the sides BC, CA and AB, respectively
...
Then,
OD · OD = OE · OE = OF · OF = R2
...

CO
y+z

...
18
...
Then we can deduce that
la
=
sin2 A

AX
AY sin2 A
ha
2R sin2 A
sin A
1
ha
since
=
a sin A
a
2R

≥
=
≥3

3

ha
hb
hc
a sin A b sin B c sin C

=3
since ha = b sin C, hb = c sin A, hc = a sin B
...
19
...
Since 1 < 2 < · · · <
n, we have, using the rearrangement inequality (1
...

Then, |A + B| = |(n + 1) (x1 + · · · + xn )| = n + 1, hence A + B = ±(n + 1)
...

2

4
...

2
2
Let y1 ,
...
, xn such that 1y1 +2y2 +· · ·+nyn =
C takes the maximum value with C ≤ − n+1
...

Since |yi |, |yi+1 | ≤ n+1 , we can deduce that D − C = yi − yi+1 ≤ n + 1; hence
2
D ≤ C + n + 1 and therefore C < D ≤ C + n + 1 ≤ n+1
...
Thus − n+1 ≤ D ≤ n+1 and then |D| ≤ n+1
...
20
...

Note that a + b − c = 2xy , b + c − a = 2yz , c + a − b = 2zx are positive
...

Solution 3
...
First, prove that a centrally symmetric hexagon ABCDEF has
opposite parallel sides
...
Now, if we reﬂect
2
the triangle P QR with respect to the symmetry center of the hexagon, we get
the points P , Q , R which form the centrally symmetric hexagon P R QP RQ ,
inscribed in ABCDEF with area 2(P QR)
...
22
...
Using the AM-GM inequality leads to 3 X1 ≥
3
4
1
inequalities hold for the other indexes and this implies that X ≥ i=1 xi
...

4
4
4
Thanks to the AM-GM inequality we get
therefore X ≥ i=1 xi
...
23
...

Solution 3
...
If α = ∠ACM and β = ∠BDM , then
α+β =

π
4
...

√ Another method uses the fact that the inequality is equivalent to (M CD) ≥
√
3 3(M AB) which is equivalent to h+l ≥ 3 3, where l is the length of the side of
h
the square and h is the length of the altitude from M to AB
...

170

Solutions to Exercises and Problems

Solution 3
...
First note that
BM and CN are less than a
...
26
...
Now, use the fact that AL,

2

, it is suﬃcient to see that

PB
PA

+

= 1
...
The triangles AP G and BP B are similar, as well as AQG and
CQC , thus P B = BB and QC = CC
...

Solution 3
...
Let Γ be the circumcircle of ABC, and let R be its radius
...
For any point P other than O, let P be its inverse
...
Since22
P Q =

R2 · P Q
OP · OQ

for two points P , Q (distinct from O) with inverses P , Q , we have
AA1
x+y+z
R2 · A A1
AA1
=
,
=
=
OA1
OA · OA1 · OA1
OA
y+z
where x, y, z denote the areas of the triangles OBC, OCA, OAB, respectively
...

=
=
OB1
z+x
OC1
x+y
Thus
AA1
BB1
CC1
+
+
= (x + y + z)
OA1
OB1
OC1
For the last inequality, see Exercise 1
...

22 See

[5, page 132] or [9, page 112]
...

2

4
...
28
...
Similarly, GN = 2(ABC)
...

Therefore

4(ABC)2 sin B
GL · GN sin B
=
2
18ac
(ABC)2 b2
4(ABC)2 b2
=
=
(18abc)(2R)
(9R abc )(4R)
4R

(GN L) =

=
Similarly, (GLM ) =

(ABC) c2
9·4R2

(ABC) b2

...

a 2 + b 2 + c2
4R2

=

Therefore,

R2 − OG2

...

For the other inequality, note that OG = 1 OH
...
Therefore,
R2 − 1 OH 2
R2 − 1 R2
4
2
(LM N )
9
9
=

...
29
...
Thus,

f (ab) + f (bc) + f (ca)
≥f
3

ab + bc + ca
3
=
3
3 + ab + bc + ca
3
1
≥
= ,
2 + b 2 + c2
3+a
2

the last inequality follows from ab + bc + ca ≤ a2 + b2 + c2
...

2 + b 2 + c2
1 + ab 1 + bc 1 + ca
3 + ab + bc + ca
3+a
2
The ﬁrst inequality follows from inequality (1
...
27
...
30
...
The desired inequality
becomes
x y
+
y
x

+

y
z
+
z
y

+

z
x
+
+3
x z

y
z
x
+ +
x y
z

≥ 15,

172

Solutions to Exercises and Problems

which can be proved using the AM-GM inequality
...

b + 2c c + 2a a + 2b
ab + 2ca bc + 2ab ca + 2bc
3(ab + bc + ca)
The inequality follows from inequality (1
...
It remains to prove the inequality
(a + b + c)2 ≥ 3(ab + bc + ca), which is a consequence of the Cauchy-Schwarz
inequality
...
31
...
11) or use the Cauchy-Schwarz inequality with
a
b
c
d
√
,√
,√
,√
a+b
b+c
c+d
d+a

√
√
√
√
and ( a + b, b + c, c + d, d + a)
...
32
...
The similarity
between the triangles ADE and ABC gives us
DE
perimeter of ADE
2x
=
=

...

x(y+z)
x+y+z

≤

x+y+z

...
33
...
Using the sine law in the triangles AE F , BCE and BDF , we
obtain
AE
AF sin F
sin F
BD
AE
sin E
=
=
=
=
,
·
EC
sin E
BC sin B
sin B
FD
EC
therefore E = E
...
Use the fact that the
triangles ECG and EAH are similar and also use Menelaus’s theorem for the
triangle CAD with transversal EF B to conclude that AH = DB
...

Solution 3
...
Note that ab + bc + ca ≤ 3abc if and only if
(a + b + c)

1 1 1
+ +
a b
c

1
a

+

1
b

+

≥ 9,

we should have that (a + b + c) ≥ 3
...

2

1
c

≤ 3
...
3 Solutions to the problems in Chapter 3

173

xi
Solution 3
...
Take yi = n−1 for all i = 1, 2,
...

1 + y1
1 + y2
1 + yn
Then

1
>
1 + yi

1−
j=i

1
1 + yj

≥ (n − 1) n−1

=
j=i

yj
1 + yj

y1 · · · yi · · · yn
ˆ
(1 + y1 ) · · · (1 + yi ) · · · (1 + yn )

,

where y1 · · · yi · · · yn is the product of the y’s except yi
...

x1
y1 ,
...
36
...
, xn yn to get

and

2

2

(x1 + · · · + xn ) =
≤
Now use the hypothesis

xn
yn

x1 √
xn √
x1 y1 + · · · +
xn yn
y1
yn
x1
xn
(x1 y1 + · · · + xn yn )
...

Solution 3
...
Since abc = 1, we have
(a − 1)(b − 1)(c − 1) = a + b + c −

1 1 1
+ +
a b
c

and similarly
(an − 1)(bn − 1)(cn − 1) = an + bn + cn −

1
1
1
+ n+ n
an
b
c

...

Solution 3
...
We prove the claim using induction on n
...

Now assuming the claim is true for n, we can prove it is true for n + 1
...
, 2n, we have
n2 + i =

n2 + i − n <

n2 + i +

i
2n

2

−n=

i

...

2

Solution 3
...
Let us prove that the converse aﬃrmation, that is, x3 + y 3 > 2,
2

2

implies that x2 + y 3 < x3 + y 4
...

3

x3 +y 3
2

Then x2 − x3 < y 3 − y 2 ≤ y 4 − y 3
...

Second solution
...

Thus, x3 + y 3 ≤ x3 + y 4 + y 2 − y 3 ≤ x2 + y 2 , since x3 + y 4 ≤ x2 + y 3
...
40
...

(x0 − x1 )
(xn−1 − xn )
√
√
√
4
5
3
Solution 3
...
Since a+3b ≥ ab3 , b+4c ≥ bc4 and c+2a ≥ ca2 , we can deduce
4
5
3
that
11 19 17
(a + 3b)(b + 4c)(c + 2a) ≥ 60a 12 b 20 c 15
...

Solution 3
...
We have an equivalence between the following inequalities:
7(ab + bc + ca) ≤ 2 + 9 abc
⇔

7(ab + bc + ca)(a + b + c) ≤ 2(a + b + c)3 + 9 abc

⇔

a2 b + a b2 + b2 c + b c2 + c2 a + c a2 ≤ 2(a3 + b3 + c3
...

Solution 3
...
Let E be the intersection of AC and BD
...

|AC − BD|
|AE − EB|
|AB|
Using the triangle inequality in ABE, we have |AE−EB| ≥ 1 and we therefore
conclude that |AB − CD| ≥ |AC − BD|
...

4
...
44
...
First, prove that the inequality is valid for j = n, that is, a1 + · · · +
an ≥ n+1 an ; use the fact that 2(a1 +· · ·+an ) = (a1 +an−1 )+· · ·+(an−1 +a1 )+2an
...

We provide another proof of a1 + · · · + aj ≥ j(j+1) an , once again using
2n
induction
...

a1 +
2
2
2
2
2
2
Now, let us suppose that the aﬃrmation is valid for n = 1,
...

...

aj
a2
+ ···+
≥ aj
...

ja1 + (j − 1) + · · · +
2
j
Adding on both sides the identity
aj
a2
= a j + · · · + a1 ,
a1 + 2 + · · · + j
2
j
we obtain
(j + 1) a1 +

aj
a2
+ ···+
2
j

Hence

≥ (a1 + aj ) + (a2 + aj−1 ) + · · · + (aj + a1 ) ≥ jaj+1
...

2
j
j+1
on both sides of the inequality provides the ﬁnal step in the
a1 +

a

j+1
Finally, adding j+1
induction proof
...

j(j + 1) 2n

176
Solution 3
...

Solutions to Exercises and Problems

⎞4

⎛

⎛

⎞2

xi ⎠ = ⎝

⎝
1≤i≤n

xi xj ⎠

x2 + 2
i

1≤i≤n

⎛
≥ 4⎝

1≤i≤n

1≤i
⎞⎛

x2 ⎠ ⎝2
i

⎛
= 8⎝

xi xj ⎠
1≤i
⎞⎛
x2 ⎠ ⎝
i

1≤i≤n

⎞

⎞

xi xj ⎠
1≤i
xi xj (x2 + · · · + x2 )
1
n

=8
1≤i
xi xj (x2 + x2 )
...

To determine when the equality occurs, note that in the last step, two of
the xi must be diﬀerent from zero and the other n − 2 equal to zero; also in the
step where the AM-GM inequality was used, the xi which are diﬀerent from zero
should in fact be equal
...

√
√
Solution 3
...
Setting 3 a = x and 3 b = y, we need to prove that (x2 + y 2 )3 ≤
2(x3 + y 3 )2 for x, y > 0
...

Adding together these two inequalities and adding x6 + y 6 to both sides, we get
x6 + y 6 + 3x2 y 2 (x2 + y 2 ) ≤ 2(x6 + y 6 + 2x3 y 3 )
...

Solution 3
...
Denote the left-hand side of the inequality as S
...

Setting α = (ax)2 , β = (by)2 , γ = (cz)2 , we obtain
a2 x2
a2 x2
α
≥
=

...
3 Solutions to the problems in Chapter 3

177

Adding together the other two similar inequalities, we get
S≥

α
β
γ
+
+
β+γ
γ+α α+β

1
2

...

Solution 3
...
If XM is a median in the triangle XY Z, then XM 2 = 1 XY 2 +
2
1
1
2
2
2 XZ − 4 Y Z , a result of using Stewart’s theorem
...
Multiplying both sides of the ﬁfth equation by 4, we ﬁnd
that the left-hand side of the desired inequality equals AB 2 + BC 2 + CD2 + DA2 +
4(AC 2 + BD2 )
...
This inequality is known as the “parallelogram inequality”
...
We expand each term in AB 2 + BC 2 + CD2 + DA2 − AC 2 −
BD2 , writing for instance
AB 2 = |a − b|2 = |a|2 − 2a · b + |b|2
and then ﬁnding that the expression equals
|a|2 + |b|2 + |c|2 + |d|2 − 2(a · b + b · c + c · d + d · a − a · c − b · d)
= |a + c − b − d|2 ≥ 0,
with equality if and only if a + c = b + d, that is, only if the quadrilateral ABCD
is a parallelogram
...
49
...
Then 1 = A+2B, B 2 = C +2xyz(x+y +z) = C +2D and x4 +y 4 +z 4 =
A2 − 2C = 4B 2 − 4B + 1 − 2C = 2C − 4B + 8D + 1
...

Now, the right-hand expression is equal to 2 + B + D
...
Using the Cauchy-Schwarz inequality,
we get A ≥ B, so that B(1 − 2B) = BA ≥ B 2
...
But C ≥ xyyz + yzzx + zxxy = D as can be deduced from
the Cauchy-Schwarz inequality
...
50
...
The inequality is equivalent to
y
z

z
x
−1+
y
y

y
x
−1+
z
z

z
y
−1+
≤1
x
x

178

Solutions to Exercises and Problems

and can be rewritten as (x + z − y)(x + y − z)(y + z − x) ≤ xyz
...
See Example 2
...
3
...
If one or three of them are negative, then the left side is negative
and the inequality is clear
...

Solution 3
...
First note that abc ≤ a + b + c implies (abc)2 ≤ (a + b + c)2 ≤
3(a2 + b2 + c2 ), where the last inequality follows from inequality (1
...

By the AM-GM inequality, a2 + b2 + c2 ≥ 3 3 (abc)2 , then (a2 + b2 + c2 )3 ≥
3
3 (abc)2
...

Solution 3
...
Using the AM-GM inequality,
(a + b)(a + c) = a(a + b + c) + bc ≥ 2

abc(a + b + c)
...
Setting x = a + b, y = a + c, z = b + c, and since a, b, c are
positive, we can deduce that x, y, z are the side lengths of a triangle XY Z
...
2
...

2
Solution 3
...
Since xi ≥ 0, then xi − 1 ≥ −1
...

Adding these inequalities together for 1 ≤ i ≤ n, gives us the result
...
54
...

abc

The left-hand side inequality is now obvious
...
2
...

4/3

a
Solution 3
...
If we prove that √a2a
≥ a4/3 +b4/3 +c4/3 , it will be clear how to
+8bc
get the result
...

Apply the AM-GM inequality to each factor of
a4/3 + b

4/3

+c

4/3

2

− a4/3

2

= b

4/3

+c

4/3

a4/3 + a4/3 + b

4/3

+c

4/3

...
3 Solutions to the problems in Chapter 3

179

1
Another method for solving this exercise is to consider the function f (x) = √x ,
3
this function is convex for x > 0 (f (x) = 4√x5 > 0)
...
Applying this
2
2
2
to x = a + 8bc, y = b + 8ca and z = c + 8ab (previously multiplying by an
appropriate factor to have the condition a + b + c = 1), we get

a
b
c
1
√
+√
+√
≥ √

...

Solution 3
...
Using the Cauchy-Schwarz inequality
x
ai = 1, bi = 1+x2 +x2i+···+x2 , we can deduce that
1

2

a2
i

ai b i ≤

i

√
x1
x2
xn
+
+ ···+
≤ n
1 + x2
1 + x2 + x2
1 + x2 + · · · + x2
n
1
1
2
1

xi
2 + · · · + x2
1 + x1
i

2

x2
i
(1 + + · · · + x2 )2
i
x2
i
≤
2 + · · · + x2 )(1 + x2 + · · · + x2 )
(1 + x1
1
i−1
i
1
1
−

...

i

b2 < 1
...

1+x2
1

2

≤1−

Adding together these inequal1
< 1
...
57
...

Therefore, we have two cases to consider
...
In this case we have a + b + c = 0, and therefore
a 3 + b 3 + c3
abc

2

2

=

a3 + b3 − (a + b)3
−ab(a + b)

=

(a + b)2 − a2 + ab − b2
ab

2

=

3ab
ab

2

= 9
...

=
ab
b
a
If a and b have the same sign, we see that this expression is not less than 5, and
its square is therefore no less than 25
...

b
b
b
Thus, the smallest possible value is 9
...
58
...
Using AM-GM inequality again gives
a+b+c
3
1
X ≤ 3 and Y ≤ 2 a+b+c , then XY ≥ 27 (a+b+c)2
...
59
...

Second solution
...

≥ (abc)
abc
In the ﬁrst two inequalities we applied inequality (1
...

Solution 3
...
Take f (x) as f (x) =

x
1−x
...
But
3

is convex
...

3
3
Solution 3
...

1
a
+
b+c 2

b
1
+
c+a 2

c
1
+
a+b 2

≥1

is equivalent to (2a + b + c)(2b + c + a)(2c + a + b) ≥ 8(b + c)(c + a)(a + b)
...

4
...
62
...
Since (a + b)2 ≥ 4ab,
c2
a2
b2
(b + c)2 ≥ 4bc and (c + a)2 ≥ 4ca, we can deduce that
(a + b)2
(b + c)2
(c + a)2
4ab 4bc 4ca
3 (ab)(bc)(ca)
+
+
≥ 2 + 2 + 2 ≥ 12
= 12
...
63
...
Adding similar
√
inequalities for y, z, we get x2 + y 2 + z 2 + 2( x + y + z) ≥ 3(x + y + z) =
(x + y + z)2 = x2 + y 2 + z 2 + 2(xy + yz + zx)
...
64
...

c +
a +
b
...

b
Solution 3
...
Since (1 − a)(1 − b)(1 − c) = 1 − (a + b + c) + ab + bc + ca − abc and
since a + b + c = 2, the inequality is equivalent to
1

...
The other one follows from the AM-GM inequality
...
66
...

A

B1

C1

B

A1

M

C

Moreover, (AA1 M ) = 3 (ABC)
...

+
+
≤
ma mb
mb mc
mc ma
4 (AA1 M )

182

Solutions to Exercises and Problems

Now, the last inequality will be true if the triangle with side-lengths a, b, c and
area S satisﬁes the following inequality:
√
1
1
1
3 3
+
+
≤

...
4
...

Solution 3
...
Substitute cd =
inequality becomes

1
ab

and da =

1
bc ,

so that the left-hand side (LHS)

1 + ab
1 + ab
1 + bc
1 + bc
+
+
+
1+a
ab + abc
1+b
bc + bcd
1
1
+
+ (1 + bc)
= (1 + ab)
1 + a ab + abc
Now, using the inequality

1
x

1
y

+

≥

4
x+y ,

(4
...

we get

4
4
+ (1 + bc)
1 + a + ab + abc
1 + b + bc + bcd
1 + bc
1 + ab
+
1 + a + ab + abc 1 + b + bc + bcd
a + abc
1 + ab
+
= 4
...
68
...

2
(b + c)
4

Using the Cauchy-Schwarz inequality leads us to
2
2
2
(la + lb + lc )2 ≤ 3(la + lb + lc )
3
≤ ((a + b)2 + (b + c)2 + (c + a)2 − a2 − b2 − c2 )
4
3
≤ (a + b + c)2
...
69
...

b+c c+a a+b
2a + b + c 2b + a + c 2c + a + b
Now, using the fact that
2

1
x

+

1
y

≥

1
1
1
+
+
b+c c+a a+b

which proves the inequality
...
3 Solutions to the problems in Chapter 3

183

Solution 3
...
We may take a ≤ b ≤ c
...

2
2
2
Since a ≤ b, we can deduce that
b+

a
2

n

= bn + nbn−1
> bn +

a
+ other positive terms
2

n n−1
ab
≥ b n + an
...

2
2

Second solution
...

Since a + b + c = 1, we can deduce that x + y + z = 1
...
Therefore
n

n

1
n

n

n

1
n

n

n

(a + b ) + (b + c ) + (c + a )

1
n

< a+b+c+

√
n

√
n
2z
...

2(x + y + z) = 1 +
2

Solution 3
...
First notice that if we restrict the sums to i < j, then they are
halved
...

Consider the sum of all the |xi − xj | with i < j
...
Thus, we get
−(n − 1)x1 − (n − 3)x2 − (n − 5)x3 − · · · + (n − 1)xn =

(2i − 1 − n)xi
...

i
Looking at the sum at the other side of the desired inequality, we immediately
see that it is n x2 − ( xi )2
...
Thus, we can ﬁnish if we can prove that (2i−1−n)2 = n(n3−1) ,
(2i − 1 − n)2 = 4

i2 − 4(n + 1)

i + n(n + 1)2

2
n(n + 1)(2n + 1) − 2n(n + 1)2 + n(n + 1)2
3
1
= n(n + 1)(2(2n + 1) − 6(n + 1) + 3(n + 1))
3
1
= n(n2 − 1)
...

Second solution
...
, xn form an arithmetic
progression, that is, when xi − xj = r(i − j) with r > 0, then consider the following
inequality which is true, as can be inferred from the Cauchy-Schwarz inequality,
⎛
⎞2
⎝

|i − j||xi − xj |⎠ ≤
i,j

(i − j)2
i,j

(xi − xj )2
...

2
2
Since (i − j)2 = (2n − 2) · 12 + (2n − 4) · 22 + · · · + 2 · (n − 1)2 = n (n6 −1) ,
i,j

we need to prove that
i,j

|i − j| |xi − xj | =

n
2

i,j

|xj − xj |
...
On the left-hand side the coeﬃcient is
(i − 1) + (i − 2) + · · · + (i − (i − 1)) − ((i + 1) − i) + ((i + 2) − i) + · · · + (n − i))
=

n(2i − n − 1)
(i − 1)i (n − i)(n − i + 1)
−
=

...

1+
− 1⎠ = ((i − 1) − (n − i)) =
2 i2
2
j>i
Since they are equal we have ﬁnished the proof
...
3 Solutions to the problems in Chapter 3

185

Solution 3
...
Let xn+1 = x1 and xn+2 = x2
...
, n}
...

i=1

The inequality is equivalent to
n
i=1

ai
n2

...

n
3
b1 · · · bn

On the other hand and using again the AM-GM inequality, we get
n
i=1

√
n
a1 · · · an
ai
a1
an
n
n2

...
73
...
Since the numbers ab, bc and ca are non-negative, we have

P (x)P

1
x

1
1
+b +c
x2
x
1
1
+ bc x +
= a2 + b2 + c2 + ab x +
x
x
= (ax2 + bx + c) a

+ ca x2 +

1
x2

≥ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2 = P (1)2
...
Consequently, for any positive real
x we have
1
P (x)P
≥ (P (1))2
x
with equality if and only if either x = 1 or b = c = 0
...
Using the Cauchy-Schwarz inequality we get
P (x)P

1
x

1
1
+b +c
2
x
x
⎛
√
√ 2
√ 2
√ 2
a
⎝
= ( ax) + ( bx) + ( c)
x
= (ax2 + bx + c) a

≥

√
√
√
a √
b √ √
+ bx √ + c c
ax
x
x

2

+

√
b
√
x

2

⎞
√ 2
+ ( c) ⎠

2

= (a + b + c)2 = (P (1))2
...
74
...

The last inequality follows after using Muirhead’s theorem
...
Use inequality (1
...

Solution 3
...
Applying the AM-GM inequality to each denominator, one obtains
1
1
1
1
1
1
+
+
≥
+
+

...
11) leads us to
1
9
1
1
(1 + 1 + 1)2
=
= 1
...
76
...

Identity (1
...

2

4
...
77
...
The points O, M , X are collinear and OCX and
OM C are similar right triangles
...

OX
OC
OA
Since OC = R = OA, we have OM = OX
...

It now suﬃces to show that OM ≤ r
...
Since ABC is an acute triangle, O and I lie inside ABC
...

2
2
2
2
Similarly, we have ∠OCM ≤ ∠ICM
...

Solution 3
...
Setting a = x2 , b = y 2 , c = z 2 , the inequality is equivalent to
x6 + y 6 + z 6 ≥ x4 yz + y 4 zx + z 4 xy
...

√ √
√
Solution 3
...
Use the Cauchy-Schwarz inequality to see that xy + z = x y +
√
√ √
√
√
√
z z ≤ x + z y + z = xy + z(x + y + z) = xy + z
...
Therefore,
√
√
√
√
√
√
xy + z + yz + x + zx + y ≥ xy + yz + zx + x + y + z
...
80
...
4
...

3

Now,
a 3 + b 3 + c3 ≥

√
a+b+c 2
3
(a + b2 + c2 ) ≥ abc(ab + bc + ca) ≥ ab + bc + ca,
3

where we have used the AM-GM and the Cauchy-Schwarz inequalities
...
81
...
4
...
On the other
hand,
4(ab + bc + ca) − 1 ≥ a2 + b2 + c2 ≥ ab + bc + ca,
therefore 3(ab + bc + ca) ≥ 1
...
Consequently, a+b+c = 1 and 3(ab+bc+ca) = (a+b+c)2 ,
which implies a = b = c = 1
...
82
...

Hence |a| + |b| + |c| ≤ 3
...
The requested inequality is then obtained
by summation
...
83
...

=
·
AA1
(ABC)
4R1
AB · AC · BC
Now, we have to prove that
OB · OC · BC + OA · OB · AB + OA · OC · AC ≥ AB · AC · BC
...

That is,
|b2 c − c2 b| + |a2 b − b2 a| + |c2 a − a2 c| ≥ |ab2 + bc2 + ca2 − a2 b − b2 c − c2 a|,
which is obvious by the triangle inequality
...
84
...
, j1 , i2 , i2 + 1,
...
, ip ,
...
, p − 1
...
Then
ai = Sjp − Sip −1 + Sjp−1 − Sip−1 −1 + · · · + Sj1 − Si1 −1
i∈S

and

(ai + · · · + aj )2 =
1≤i≤j≤n

(Si − Sj )2
...
9)

i=1

because this means neglecting the same non-negative terms on the right-hand side
of the inequality
...
9) reduces to
p

xi xj ≤ (p − 1)

4
1≤i≤j≤p
j−i even

i=1

x2
...
3 Solutions to the problems in Chapter 3

189

This can be obtained adding together inequalities with the form 4xi xj ≤ 2(x2 +x2 ),
i
j
i < j, j − i =even (for odd i, xi appears in such inequality p−1 times, and for
2
even i, xi appears in such inequality p − 1 times)
...
85
...
Then a2 +b2 +c2 = x2 −2y,
a2 b2 +b2 c2 +c2 a2 = y 2 −2xz, a2 b2 c2 = z 2 , and the inequality to be proved becomes
z 2 + 2(y 2 − 2xz) + 4(x2 − 2y) + 8 ≥ 9y or z 2 + 2y 2 − 4xz + 4x2 − 17y + 8 ≥ 0
...

Also,
a2 b2 + b2 c2 + c2 a2 = (ab)2 + (bc)2 + (ca)2
≥ ab · ac + bc · ab + ac · bc
= (a + b + c)abc = xz,
and thus y 2 = a2 b2 + b2 c2 + c2 a2 + 2xz ≥ 3xz
...

Second solution
...

Since 3(a2 + b2 + c2 ) ≥ 3(ab + bc+ ca) and 2(a2 b2 + b2 c2 + c2 a2 )+ 6 ≥ 4(ab + bc+ ca)
√
(for instance, 2a2 b2 + 2 ≥ 4 a2 b2 = 4ab), it is enough to prove that
(abc)2 + a2 + b2 + c2 + 2 ≥ 2(ab + bc + ca)
...
90 tells us that it is enough to prove that (abc) + 2 ≥
3
2 b2 c2 , but this follows from the AM-GM inequality
...
86
...

abc

190

Solutions to Exercises and Problems

It is easy to see that 3((ab)2 + (bc)2 + (ac)2 ) ≥ (ab + bc + ac)2 (use the CauchySchwarz inequality)
...

3

3
1
≤
,
abc
abc

But this last inequality follows from the AM-

(abc)2 = (ab)(bc)(ca) ≤

ab + bc + ca
3

The equality holds if and only if a = b = c =

3

=

1

...

3

Solution 3
...
Using symmetry, it suﬃces to prove that t1 < t2 + t3
...

Using the AM-GM inequality we get
√
2
ti
tj
, t2 + t3 ≥ 2 t2 t3 and
+
≥ 2 for all i, j
...

ti
≥ n + 2√
+2
+2
n2 + 1 >
ti
t1
2
a
t2 t3
i=1
i=1
√
√
2
Hence 2a + a − 5 < 0, which implies 1/2 < a = t1 / t2 t3 < 2
...

1
1
+
t2
t3

≥√

Solution 3
...
Note that 1 + b − c = a + b + c + b − c = a + 2b ≥ 0
...

3
3
Similarly,
√
bc − ba
b 31+c−a≤b+
3
√
ca − cb
3

...

4
...
89
...

Therefore, it can only be 2 or 4 negative numbers between the numbers
in the inequality
...
If yi = |xi |,
2
2
2
then it is clear that y1 + y2 + · · · + y6 = 6, y1 + y2 = y3 + · · · + y6 and that
x1 x2 · · · x6 = y1 y2 · · · y6
...

Also, the AM-GM inequality yields
y3 y4 y5 y6 ≤

4

y3 + y4 + y5 + y6
4

=

4

y1 + y2
4

=

1 4
A
...

On the other hand, the Cauchy-Schwarz inequality implies that
2
2
2(y1 + y2 ) ≥ (y1 + y2 )2 = 4A2
2
2
2
2
4(y3 + y4 + y5 + y6 ) ≥ (y3 + y4 + y5 + y6 )2 = 4A2
...

1
6
24 A

≤

b c
Solution 3
...
Use the Cauchy-Schwarz inequality with (1, 1, 1) and ( a , c , a ) to
b
obtain
2
c
b2
c2
a b
a2
(12 + 12 + 12 )
+ +
+ 2+ 2 ≥

...

= 3
...

2
b
c
a
c
a b
b
c a
Adding

a
c

+

b
a

+

c
b

to both sides yields the result
...
91
...

After substituting in the given inequality, we need to prove that
(a2

b2
c2
1
a2
+ 2
+ 2
≥
...

Now, 3(xy + yz + zx) + 6(x + y + z) ≥ xyz + 2(xy + yz + zx) + 4(x + y + z) + 8
if and only if xy + yz + zx + 2(x + y + z) ≥ xyz + 8 = 72, but using the AM-GM
inequality leads to x + y + z ≥ 12 and xy + yz + zx ≥ 48, which ﬁnishes the proof
...
92
...

− 3 2
x5 + y 2 + z 2
x (x + y 2 + z 2 )
x (x + y 2 + z 2 )(x2 + y 2 + z 2 )
Then
x5 − x2
≥
x5 + y 2 + z 2

x5 − x2
x3 (x2 + y 2 + z 2 )
1
1
x2 −
= 2
2 + z2
x +y
x
1
≥ 2
(x2 − yz) ≥ 0
...
8)
...
The

Second solution
...

x5 + y 2 + z 2
x5 + y 2 + z 2
x + y2 + z 2
Now we need to prove that
1
1
1
3
+ 5
+ 5
≤ 2

...
3 Solutions to the problems in Chapter 3

193

Using the Cauchy-Schwarz inequality we get
(x2 + y 2 + z 2 )2 ≤ (x2 · x3 + y 2 + z 2 )(x2 ·
and since xyz ≥ 1, then x2 ·

1
x3

=

1
x

1
+ y2 + z 2)
x3

≤ yz, and we have that

(x2 + y 2 + z 2 )2 ≤ (x5 + y 2 + z 2 )(yz + y 2 + z 2 )
therefore
yz + y 2 + z 2
≤
(x2 + y 2 + z 2 )2
3
= 2

...
93
...

a(b + 1)
(b + 1)
b(c + 1)
(c + 1)
c(a + 1)
(a + 1)

The last inequality follows after using the AM-GM inequality for six numbers
...
94
...
Since ∠BOC =
2∠A, ∠COA = 2∠B and ∠AOB = 2∠C, we have that
R2
(sin 2A + sin 2B + sin 2C)
2
2A + 2B + 2C
R2
3 sin
≤
2
3
√
2
R
2π
3 3R2
=
3 sin
=

...

On the other hand, since BOC is isosceles, the perpendicular bisector OA of
BC is also the internal bisector of the angle ∠BOC, so that ∠BOA = ∠COA =
∠A; similarly ∠COB = ∠AOB = ∠B and ∠AOC = ∠BOC = ∠C
...

2
2

194

Solutions to Exercises and Problems
2

Therefore, the area of the triangle B OC is (B OC ) = R (tan B + tan C)
...
Then,
8
8
R2
(tan A + tan B + tan C)
4
A+B+C
R2
3 tan
≥
4
3
√
2
R
π
3 3R2
=
3 tan
=

...

2
Hence,
√
3 3R2
≥ (ABC)
...
95
...

Now, using the Cauchy-Schwarz inequality in the following way
1
1
1
√ √ +√ √ +√ √
ab ca
bc ab
ca bc

2

≤

1
1
1
+
+
ab bc ca

1
1
1
+
+
ca ab bc

,

the result follows
...
96
...

i=1

i=j

On the other hand,
2

n

ai aj =
i=j

ai
i=1

n

−

a2
i

= A2 − A
...
97
...
Let dk = ak+1 − ak
for k = 1,
...
Then d = d1 + · · · + dn−1
...
Then,

4
...

=
k=1

Since k(n − k) ≥ (n − 1) (because (k − 1)(n − k − 1) ≥ 0) and 4k(n − k) ≤ n2
2
(from the AM-GM inequality), we obtain (n − 1)d ≤ s ≤ n4 d
...

For the second equality notice that 4k(n − k) = n2 ⇔ k = n − k
...
If n is even, say n = 2k, then only dk can be diﬀerent from
zero and then a1 = · · · = ak ≤ ak+1 = · · · = a2k
...
98
...

This satisﬁes the identities P (b) = P (c) = P (−b − c) = 0, therefore P (t) =
(b − c)(t − b)(t − c)(t + b + c), since the coeﬃcient of t3 is (b − c)
...

The problem is to ﬁnd the least number M such that the following inequality holds
for all numbers a, b, c:
|(a − c)(a − b)(b − c)(a + b + c)| ≤ M (a2 + b2 + c2 )2
...
Therefore, we can assume, without loss of generality, that a2 + b2 + c2 =
1
...

Note that
[3(a2 + b2 + c2 )]2 = [2(a − b)2 + 2(a − c)(b − c) + (a + b + c)2 ]2
≥ 8 |(a − c)(b − c)| [2(a − b)2 + (a + b + c)2 ]
√
≥ 16 2 |(a − c)(b − c)(a − b)(a + b + c)|
√
= 16 2P
The two inequalities are obtained using the AM-GM inequality
...

6 2

9
√
16 2

because the equality holds

Solution 3
...
For a = 2, b = c = 1 and n ≥ 3, the inequality is not true
...

3
For the case n = 2, let x = ab+bc+ca; now since a2 +b2 +c2 = (a+b+c)2 −2(ab+bc+
ca) = 9−2x and x2 = (ab+bc+ca)2 ≥ 3(a2 bc+ab2 c+abc2 ) = 3abc(a+b+c) = 9abc,
the inequality is equivalent to abc(9 − 2x) ≤ 3, but it will be enough to prove that
x2 (9 − 2x) ≤ 27
...

√
3
Solution 3
...
First, the AM-GM inequality leads us to ca + c + a ≥ 3 c2 a2
...

≥
3 2 2
ca + c + a + 1
(c + 1)(a + 1)
(c + 1)
3 c a +1
Similarly for the other two terms of the sum; therefore
(a + 1)(b + 1)2
(b + 1)(c + 1)2
(c + 1)(a + 1)2
√
√
√
+
+
3 2 2
3
3
3 c a +1
3 a2 b 2 + 1
3 b 2 c2 + 1
(c + 1)2
(a + 1)2
(b + 1)2
+
+

...
11)
...
101
...
Moreover,
2
3
8
(a2 + b2 + c2 )(a + b + c) + 4abc
4abc
=
3
3
2
2
2
2((y + z) + (z + x) + (x + y) )(x + y + z) + 4(y + z)(z + x)(x + y)
=
3
4
= ((x + y + z)3 − xyz)
3
3
3
13
1
4
=

...

Solution 3
...
Apply Ravi’s transformation a = y + z, b = z + x, c = x + y, so
that the inequality can be rewritten as
(2x)4
(2y)4
(2z)4
+
+
(z + x)(2x) (x + y)(2y) (y + z)(2z)
≥ (y + z)(z + x) + (z + x)(x + y) + (x + y)(y + z)
...
3 Solutions to the problems in Chapter 3

197

From inequality (1
...
27, we obtain
(2x)4
(2y)4
8(x2 + y 2 + z 2 )2
(2z)4
+
+
≥ 2
(z + x)(2x) (x + y)(2y) (y + z)(2z)
x + y 2 + z 2 + xy + yz + zx
8(x2 + y 2 + z 2 )2

...

a+b
Solution 3
...
The substitution x = a−b , y = b+c , z = c+a has the property
b−c
c−a
that xy + yz + zx = 1
...

Solution 3
...
It will be enough to consider the case x ≤ y ≤ z
...

On the one hand, we have xz = 1 − xy − yz = 1 − (y − a)y − y(y + b) =
1 − 2y 2 + ay − by and on the other, xz = (y − a)(y + b) = y 2 − ay + by − ab
...

If 2xz = 1, then y = 0 and xz = 1, a contradiction, therefore xz < 1
...
However, xz = 2n (n − n ) = 1 − 2n2 can be as close as we wish to 1 , therefore,
2
2
1
the value 2 cannot be improved
...
105
...
Then, the inequality is
equivalent to
a + 2r
a
−
a
a + 2r

+

r
2a + r
−
r
2a + r

>

9

...

2

≥ 2, it is enough to prove that
a
r
3
+
<
...

a + 2r 2a + r
a+r a+r
2

198

Solutions to Exercises and Problems

Solution 3
...
Inequality (1
...
Thus, it is enough to prove that a + b + c ≥
3
Since (x + y + z)2 ≥ 3(xy + yz + zx), we have

1 1 1
+ +
a b
c

(a + b + c)2 ≥

2

≥3

1
1
1
+
+
ab bc ca

=

3
abc
...

Solution 3
...
By means of the Cauchy-Schwarz inequality we get
(a + b + 1)(a + b + c2 ) ≥ (a + b + c)2
...

≥
a+b+1 b+c+1 c+a+1
Therefore,
2(a + b + c) + (a2 + b2 + c2 ) ≥ (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca),
and the result follows
...
108
...
Let S be the intersection
of BP with the tangent to the circle at Q
...

On the other hand, BS + SC ≥ BQ + QC, then BP + P C is minimum if
P = Q
...
Since the triangle AM Q is isosceles and M T
is one of its altitudes, then M T = ZQ where Z is the foot of the altitude of Q
over AB
...
By symmetry, BQ
should be also an altitude and then P is the orthocenter
...
109
...
Since the triangle M N P
is acute, H belongs to the interior of the triangle M N P ; hence, it belongs to the
interior of the triangle ABC too, and therefore
x ≤ HA + HB + HC ≤ HM + HN + HP ≤ 2X
...
3 Solutions to the problems in Chapter 3

199

The second inequality is evident, the other two will be presented as the following
two lemmas
...
If H is an interior point or belongs to the sides of a triangle ABC,
and if A , B , C are its projections on BC, CA, AB, respectively, then x ≤
HA + HB + HC , where x is the length of the shortest altitude of ABC
...

HA
HB
HC
(BHC) (CHA) (AHB)
HA + HB + HC
≥
+
+
= 1
...
If M N P is an acute triangle and H is its orthocenter, then HM +HN +
HP ≤ 2X, where X is the length of the largest altitude of the triangle M N P
...
Suppose that ∠M ≤ ∠N ≤ ∠P , then N P ≤ P M ≤ M N and so it happens
that X is equal to the altitude M M
...

Let H be the symmetric point of H with respect to N P ; since M N H P is
a cyclic quadrilateral, Ptolemy’s theorem tells us that
H M · N P = H N · M P + H P · M N ≥ H N · N P + H P · N P,
and then we get H N + H P ≤ H M = HM + 2HM
...
110
...
Then
x + y ≤ z + x ≤ y + z, xy ≤ zx ≤ yz, 2z 2 (x + y) ≥ 2y 2 (z + x) ≥ 2x2 (y + z),
√ 21
≤ √ 21
≤ √ 21

...

to both sides of the last inequality, we obtain

2x2 + 2yz
2x2 (y + z)

≥

2x2 + xy + zx
2x2 (y + z)
2

=
≥

2x + x(y + z)
2x2 (y + z)
2

2x3 (y + z)

2x2 (y + z)
√
√
√
= 2( x + y + z) = 2
...
First, note that
x2 + yz
2x2 (y

+ z)

=

x2 − x(y + z) + yz
2x2 (y

+ z)

(x − y)(x − z)

x(y + z)

+

2x2 (y + z)

y+z
2
+ z)
√
√
y+ z
(x − y)(x − z)

...

Then, it is enough to prove that
(x − y)(x − z)
2x2 (y + z)

+

(y − z)(y − x)
2y 2 (z + x)

+

(z − x)(z − y)
2z 2 (x + y)

≥ 0
...
Then (x−y)(x−z) ≥ 0, and
2x (y+z)

(y − z)(y − x)
2y 2 (z

+

(z − x)(z − y)

+ x)
2z 2 (x + y)
(x − z)(y − z) (y − z)(x − y)
(x − y)(y − z) (y − z)(x − y)
=
−
≥
−
2z 2 (x + y)
2y 2 (z + x)
2z 2 (x + y)
2y 2 (z + x)
= (y − z)(x − y)

1
2z 2 (x

+ y)

−

1
2y 2 (z

+ x)

≥ 0
...

Solution 3
...
Inequality (1
...

2 + b + c2
2 + c + a2
2 + a + b2
6 + a + b + c + a 2 + b 2 + c2
Then, we need to prove that 6+a+b+c+a2 +b2 +c2 ≤ 12, but since a2 +b2 +c2 = 3,
it is enough to prove that a + b + c ≤ 3
...

The equality holds if and only if a = b = c = 1
...
112
...

a + bc
1 − b − c + bc
(1 − b)(1 − c)
(c + a)(a + b)

4
...

(c + a)(a + b) (a + b)(b + c) (b + c)(c + a)
2
This last inequality can be simpliﬁed to
4 [bc(b + c) + ca(c + a) + ab(a + b)] ≥ 3(a + b)(b + c)(c + a),
which in turn is equivalent to the inequality
ab + bc + ca ≥ 9abc
...

b
c
√
√ 2
√
√
Solution 3
...
Notice that (x y + y z + z x) = x2 y + y 2 z + z 2 x + 2(xy yz +
√
√
yz zx + zx xy)
...

Since (x + y)(y + z)(z + x) = x2 y + y 2 z + z 2 x + xy 2 + yz 2 + zx2 + 2xyz, we obtain
√
√
√
(x y + y z + z x)2 ≤ (x + y)(y + z)(z + x) + xyz
1
≤ (x + y)(y + z)(z + x) + (x + y)(y + z)(z + x)
8
9
= (x + y)(y + z)(z + x)
...
When x = y = z, the equality holds with
8
3
K = 2√2 , hence this is the minimum value
...
Apply the Cauchy-Schwarz inequality in the following way:
√
√
√ √
√ √
√ √
√
x y + y z + z x = x xy + y yz + z zx ≤ (x + y + z)(xy + yz + zx)
...

2
2
2

202

Solutions to Exercises and Problems

Solution 3
...
The left-hand side of the inequality can be written as
a2 b2 cd+ab2 c2 d+abc2 d2 +a2 bcd2 +a2 bc2 d+ab2 cd2 = abcd(ab+bc+cd+ac+ad+bd)
...
To see that the factor (ab + bc + cd + ac + ad + bd) is less than 3 we
2
can proceed in two forms
...

The second way consists in applying the AM-GM inequality as follows:
(ab + bc + cd + ac + ad + bd)
≤

a2 + b 2
b 2 + c2
c2 + d2
+
+
2
2
2

+

a 2 + c2
a2 + d2
b2 + d2
3
+
+
=
...
115
...

Now, the AM-GM inequality implies that
√
(x + y + z) ≥ 3 3 xyz ≥ 3,
(xy + yz + zx) ≥ 3 3 x2 y 2 z 2 ≥ 3,
(x2 + y 2 + z 2 ) ≥ 3 3 x2 y 2 z 2 ≥ 3
...

The equality holds when x = y = z = 1
...

But u ≥ 3 implies that (u − 2)2 ≥ 1, then (u − 2)2 + 2v ≥ 1 + 6 = 7
...

Solution 3
...
Notice that
1
1
1
1
=
=
=

...
3 Solutions to the problems in Chapter 3
The AM-GM inequality implies that 1 =

203

ab+bc+ca
3

≥

√
3
a2 b2 c2 , then abc ≤ 1
...

1 + a2 (b + c)
3a + 1 − abc
3a
Similarly,

1
1+b2 (c+a)

1
1+

a2 (b

+ c)

≤
+

1
3b

and

1
1+c2 (a+b)

1
1+

b2 (c

+ a)

+

≤

1
3c
...

=
3abc
abc
≤

Solution 3
...
The inequality is equivalent to
(a + b + c)

1
1
1
+
+
a+b b+c c+a

≥ k + (a + b + c)k = (a + b + c + 1)k
...

=
(a + b + c)2 − abc
Hence
(a + b + c)
(a + b + c + 1)

1
1
1
+
+
a+b b+c c+a

=

(a + b + c)2
≥ 1,
(a + b + c)2 − abc

and since the equality holds if and only if abc = 0, we can conclude that k = 1 is
the maximum value
...
118
...

By the Sch¨ r inequality with n = 1, Exercise 1
...

Adding these last inequalities, we get the result
...
119
...
If a, b > 0, then

1
(a−b)2

Proof
...

1
4
+ a2 + b1 − ab =
2

(a2 +b2 −3ab)2
a2 b2 (a−b)2
...

+
+
≥
(x − y)2
(y − z)2
(z − x)2
(x − z)(y − z)
Now, it is left to prove that xy + yz + zx ≥ (x − z)(y − z); but this is equivalent
to 2z(y + x) ≥ z 2 , which is evident
...
120
...

First form
...

(x − 1)2
(y − 1)2
(z − 1)2
(x − 1)2 (y − 1)2 (z − 1)2
y
x
z
Second form
...
With the previous
identities we can obtain

a2 + b2 + c2 = (a + b + c)2 − 2(ab + bc + ca)
= (a + b + c)2 − 2(a + b + c − 1)
= (a + b + c − 1)2 + 1,
therefore

a2 + b2 + c2 = (a + b + c − 1)2 + 1
...
For instance, if
we used the second form, the equality holds when a2 +b2 +c2 = 1 and a+b+c = 1
...
From
the equations we can cancel out one variable, for instance c (and since c = 1−a−b,
if we ﬁnd that a and b are rational numbers, then c will be a rational number too),
to obtain a2 + b2 + ab − a − b = 0, an identity √ we can think of as a quadratic
that
1−a±

(1−a)(1+3a)

equation in the variable b with roots b =
, which will be rational
2
k
numbers if (1 − a) and (1 + 3a) are squares of rational numbers
...
Thus, the rational numbers a = m ,
2
b = m−k+k −1 and c = 1 − a− b, when k varies in the integer numbers, are rational
2m
numbers where the equality holds
...

Notation
We use the following standard notation:

N
R
R+
⇔
⇒
a∈A
A⊂B
|x|
{x}
[x]
[a, b]
(a, b)
f : [a, b] → R
f (x)
f (x)
det A
n
i=1 ai
n
i=1 ai
i=j ai
max{a, b,
...
}
√
√x
n
x
exp x = ex
cyclic f (a, b,
...
, an except aj
the maximum value between a, b,
...

the square root of the positive real number x
the n-th root of the real number x
the exponential function
represents the sum of the function f evaluated
in all cyclic permutations of the variables a, b,
...
, xn )

(b) ≺ (a)
[b] ≤ [a]

the sum of the n! terms obtained from evaluating F in
all possible permutations of (x1 ,
...

n
n
1 2
1 2
n!
!

!

We use the following geometric notation:
A, B, C
a, b, c
A ,B ,C
∠ABC
∠A
(ABC)
(ABCD
...

the lengths of the medians of the triangle ABC
the lengths of the altitudes of the triangle ABC
the lengths of the internal bisectors of the triangle ABC
the semiperimeter of the triangle ABC
the inradius of the triangle ABC, the radius of the incircle
the circumradius of the triangle ABC, the radius of the
circumcircle
the incenter, circumcenter, orthocenter and centroid
of the triangle ABC
the centers of the excircles of the triangle ABC
...

Bibliography
[1] Altshiller, N
...
Barnes and Noble, 1962
...
, Feng, Z
...

Mathematical Olympiads 1999-2000
...

[3] Andreescu, T
...
, Lee, G
...
Mathematical Olympiads 2000-2001
...

[4] Andreescu, T
...
, Mathematical Olympiad Treasures
...

[5] Barbeau, E
...
, Shawyer, B
...
R
...
A Taste of Mathematics, vol
...

[6] Bulajich, R
...
A
...
Cuadernos de Olimpiadas de
o
ı
Matem´ticas, Instituto de Matem´ticas, UNAM, 2002
...
, G´mez Ortega, J
...
, Geometr´a
...

o
ı
Cuadernos de Olimpiadas de Matem´ticas, Instituto de Matem´ticas, UNAM,
a
a
2002
...
, Robbins, H
...

o

Fondo de Cultura

[9] Coxeter, H
...
, Geometry Revisited
...
Library, MAA,
1967
...
, 100 Great Problems of Elementary Mathematics
...

[11] Engel, A
...
Springer-Verlag, 1998
...
, Genkin, S
...
, Mathematical Circles
...
7
...

Mathematical

[13] Hardy, G
...
, Littlewood, J
...
, P`lya, G
...
Cambridge at the
o
University Press, 1967
...
, Episodes in Nineteenth and Twentieth Century Euclidean
Geometry
...
Library, MAA, 1995
...
, Geometric Inequalities
...
Library, MAA
...

[16] Larson, L
...
Springer-Verlag, 1990
...
, Elementary Inequalities
...
, Groningen, 1964
...
, Maxima and Minima Without Calculus
...

The Dolciani Math
...
, Problemas de Geometr´a
...

ı
[20] Soulami, T
...
Ellipses, 1999
...
, Calculus
...

Index
Absolute value, 2
Concavity
Geometric interpretation, 25
Convexity
Geometric interpretation, 25
discrepancy, 46
Erd˝s-Mordell
o
theorem, 81–84, 88
Euler
theorem, 66
Fermat
point, 90, 92
Function
concave, 23
convex, 20
quadratic, 4
Greater than, 1
Inequality
arithmetic mean–geometric
mean, 9, 47
weighted, 27
Bernoulli, 31
Cauchy-Schwarz, 15, 35
Engel form, 35
Euler, 67
H¨lder, 27
o
generalized, 32
harmonic mean–geometric
mean, 8
helpful, 34

Jensen, 21
Leibniz, 69
Minkowski, 28
Nesbitt, 16, 37, 65
Popoviciu, 32
power mean, 32
Ptolemy, 53
quadratic mean–arithmetic
mean, 19, 36
rearrangement, 13
Schur, 31
Tchebyshev, 18
triangle, 3
general form, 3
Young, 27
Leibniz
theorem, 68
Mean
arithmetic, 7, 9, 19, 31
geometric, 7, 9, 19, 31
harmonic, 8, 19
power, 32
quadratic, 19
Muirhead
theorem, 43, 44
Ortic
triangle, 95, 98
Pappus
theorem, 80
Pedal
triangle, 99

210
Problem
Fagnano, 88, 94
Fermat-Steiner, 88
Heron, 92
with a circle, 93
Pompeiu, 53
Real line, 1
Smaller than, 1
Smaller than or equal to, 2
Solution
Fagnano problem
Fej´r L
...
, 96
Fermat-Steiner problem
Hofmann-Gallai, 91
Steiner, 92, 94
Torricelli, 88, 90
Transformation
Ravi, 55, 73
Viviani
lemma, 88, 90

Index

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