Problem Books in Mathematics

Hayk Sedrakyan · Nairi Sedrakyan

Algebraic
Inequalities

Problem Books in Mathematics
Series Editor:
Peter Winkler
Department of Mathematics
Dartmouth College
Hanover, NH 03755
USA

More information about this series at http://www.springer.com/series/714

Hayk Sedrakyan Nairi Sedrakyan
•

Algebraic Inequalities

123

Hayk Sedrakyan
University Pierre and Marie Curie
Paris, France

Nairi Sedrakyan
Yerevan, Armenia

ISSN 0941-3502
ISSN 2197-8506 (electronic)
Problem Books in Mathematics
ISBN 978-3-319-77835-8
ISBN 978-3-319-77836-5 (eBook)
https://doi.org/10.1007/978-3-319-77836-5
Library of Congress Control Number: 2018934928
Mathematics Subject Classification (2010): 97U40, 00A07, 26D05
© Springer International Publishing AG, part of Springer Nature 2018
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To Margarita,
a wonderful wife and a loving mother
To Ani,
a wonderful daughter and a loving sister

Preface

In mathematics one often deals with inequalities. This book is designed to teach the
reader new and classical techniques for proving algebraic inequalities. Moreover,
each chapter of the book provides a technique for proving a certain type of
inequality.
The book includes techniques of using the relationship between the arithmetic,
geometric, harmonic, and quadratic means, the principle of mathematical induction,
the change of variable(s) method, techniques using the Cauchy–Bunyakovsky–
Schwarz inequality, Jensen's inequality, and Chebyshev's properties of functions,
among others. The main idea behind of the proof techniques discussed in this book
is making the complicated simple, so that even a beginner can understand complicated inequalities, their proofs and applications. This approach makes it possible
not only to prove a large variety of inequalities, but also to solve problems related to
inequalities. To explain each technique of proof, we provide examples and problems with complete proofs or hints. At the end of each chapter there are problems
for independent study. In Chapter 14 (Miscellaneous Inequalities) are included
inequalities whose proofs employ various techniques not covered in the preceding
chapters. In some cases, the proofs of Chapter 14 use several proof techniques from
the preceding chapters simultaneously. One hundred selected inequalities and their
hints are also provided in the end of Chapter 14, and interested readers are
encouraged to choose and provide any methods of proofs they prefer. In each
chapter we have tried to include inequalities belonging to the same topic and to
present them in order of increasing difficulty, using principles similar to those in
[11]. This allows the reader to try to prove these inequalities step by step and to
refer to the provided proofs only when difficulties arise. We recommend to use the
proofs provided in the book, paying more attention to the choice of the mathematical proof technique.
Most of the inequalities in this book were created by the authors. Nevertheless,
some of the inequalities were proposed in different mathematical olympiads in
different countries or have been published elsewhere (including author-created
inequalities). However, the provided solutions are different from the original ones.
Most such inequalities are included in the books [2, 5, 6, 7, 8, 9], and since the
vii

viii

Preface

name of the author of individual inequalities is unknown to us, we cite these books
as the main references. However, for well-known inequalities we have tried to
provide the name of the authors. This book was published in Seoul in Korean [13,
14] and is based on [16], which was later published in Moscow in Russian [15]. The
historical origins provided at the beginning of some chapters are mostly based on
[10] or our personal knowledge.
It was considered appropriate to give the proofs of each chapter at the end of the
same chapter.
Paris, France
Yerevan, Armenia

Hayk Sedrakyan
Nairi Sedrakyan

Contents

1

Basic Inequalities and Their Applications . . . . . . . . . . . . . . . . . . . .

1

2

Sturm's Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

9

3

The HM-GM-AM-QM Inequalities . . . . . . . . . . . . . . . . . . . . . . . . .

21

4

The Cauchy–Bunyakovsky–Schwarz Inequality . . . . . . . . . . . . . . .

45

5

Change of Variables Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

59

6

Using Symmetry and Homogeneity . . . . . . . . . . . . . . . . . . . . . . . . .

71

7

The Principle of Mathematical Induction . . . . . . . . . . . . . . . . . . . .

81

8

A Useful Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

9

Using Derivatives and Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

10 Using Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
11 Jensen's Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
12 Inequalities of Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177
13 Algebraic Inequalities in Number Theory . . . . . . . . . . . . . . . . . . . . 187
14 Miscellaneous Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197
Appendix—Power Sums Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243

ix

About the Authors

Hayk Sedrakyan is an IMO medal winner, a professor of mathematics in Paris,
France, and a professional math olympiad coach in the greater Boston area,
Massachusetts, USA. He received his doctorate in mathematics at the Université Pierre
et Marie Curie, Paris, France. Hayk is a Doctor of Mathematical Sciences in USA,
France, Armenia. He has been awarded master's degrees in mathematics from
Germany, Austria, and Armenia and completed part of his doctoral studies in Italy.
Hayk has authored several books on the topic of problem-solving and olympiad-style
mathematics published in USA and South Korea.
Nairi Sedrakyan has long been involved in national and international mathematical
olympiads, having served as an International Mathematical Olympiad (IMO) problem
selection committee member and the president of Armenian Mathematics Olympiads.
He is the author of one of the hardest problems ever proposed in the history of
International Mathematical Olympiads (the fifth problem of the 37th IMO). He has
been the leader of the Armenian IMO Team, jury member of the IMO, jury member
and problem selection committee member of the Zhautykov International Mathematical
Olympiad (ZIMO), jury member and problem selection committee member of the
International Olympiad of Metropolises, and the president of the International
Mathematical Olympiad Tournament of the Towns in Armenia. He is also the author of
a large number of problems proposed in these olympiads and has authored several
books on the topic of problem-solving and olympiad-style mathematics published in
United States, Russia, Armenia, and South Korea. The students of Nairi Sedrakyan
have obtained 20 medals (1 gold medal, 4 silver medals, 15 bronze medals) in IMO.
For his outstanding teaching, Nairi Sedrakyan received the title of best teacher of the
Republic of Armenia and was awarded special recognition by the prime minister.

xi

Overview

This book is designed to teach the reader new and classical mathematical proof
techniques for proving inequalities, in particular, to prove algebraic inequalities.
These proof techniques and methods are applied to prove inequalities of various
types. The main idea behind this book and the proof techniques discussed is making
the complicated simple, so that even a beginner can understand complicated
inequalities, their proofs and applications. The book Algebraic Inequalities is also
devoted to the topic of inequalities and can be considered a continuation of the
book Geometric Inequalities: Methods of proving [12].
It can serve teachers, high-school students, and mathematical competitors.

xiii

Chapter 1

Basic Inequalities and Their Applications

Historical origins. According to [3], mathematical inequalities first were expressed
verbally. Later, the following symbols were introduced:
Less than and greater than: The mathematical symbols < and > first appear
in the book Artis Analyticae Praxis ad Aequationes Algebraicas Resolvendas (The
Analytical Arts Applied to Solving Algebraic Equations) posthumously published
in 1631 and written by the English astronomer and mathematician Thomas Harriot
(c. 1560–1621), who was born in Oxford, England, and died in London, England.
Harriot initially used triangular symbols, but the editor modified them slightly, so that
they resemble the modern less than and greater than symbols. In his book is stated
the following: “The mark of the majority (signum majoritatis) as a > b, signifies a
greater than b and the mark of the minority (signum minoritatis) to a < b signifies
a lesser than b.”
Less than or equal to and greater than or equal to: The double-bar style of the
less than or equal to < and greater than or equal to > symbols was first employed
in 1734 by the French mathematician, geophysicist, and astronomer Pierre Bouger
(1698–1758) born in Le Croisic, France, and died in Paris, France.
In 1670, a similar reduced single-bar notation was employed by the English mathematician John Wallis. Wallis used a single horizontal bar above instead of below
the inequality symbols, leading to < and >. Wallis also introduced the symbol for
infinity ∞. His academic advisor was the English mathematician William Oughtred
(1574–1660), who introduced the multiplication symbol × and the abbreviations sin
and cos for the sine and cosine. Oughtred was born in Eton, England, and died in
Albury, England.
Not equal to, not greater than, not less than: These symbols were employed by
the Swiss mathematician, physicist, and astronomer Leonhard Euler (1707–1783),
who was born in Basel, Switzerland, and died in Saint Petersburg, Russian Empire
(now Russia). Euler introduced many modern mathematical notations, for example
the notation for a mathematical function. He is considered one of the greatest mathematicians of all time. His doctoral advisor was the prominent Swiss mathematician
Johann Bernoulli (1667–1748), who was born and died in Basel, Switzerland. Euler
© Springer International Publishing AG, part of Springer Nature 2018
H. Sedrakyan and N. Sedrakyan, Algebraic Inequalities, Problem Books
in Mathematics, https://doi.org/10.1007/978-3-319-77836-5_1

1

2

1 Basic Inequalities and Their Applications

was the doctoral advisor of six students, including the well-known Italian mathematician Joseph Louis Lagrange (1736–1813). He was born Giuseppe Lodovico
Lagrangia in Turin, Kingdom of Sardinia (now Turin, Italy) and died in Paris, France,
and his son Johann Euler (1734–1800) born and died in Saint Petersburg, Russia.
In this chapter, basic but very useful inequalities are presented. We recommend
that the reader pay special attention to these inequalities, which will lay a foundation
for more challenging inequalities in the forthcoming sections. The main idea behind
the proof techniques and methods presented here is making the complex simple, so that
even a beginner can understand complex inequalities, their proofs and applications.
A background in high-school-level algebra is sufficient for solving the inequalities
presented below.

Problems
Prove the following inequalities (1.1–1.26).
1.1. a 2 + b2 √
≥ 2ab.
1.2. a +2 b ≥ ab, where a ≥ 0, b ≥ 0.
√
ab ≥ 1 +2 1 , where a > 0, b > 0.
1.3.
a b

a 2 + b2
a+b
1.4.
≥
.
2
2
1.5.
1.6.
1.7.
1.8.
1.9.

2
, where a > 0, b > 0.
+ b1


a 2 + b2
a+b 2
≥ 2 .
2
a + b > 1 + ab, where b < 1 < a.
a 2 + b2 > c2 + (a + b − c)2 , where b < c < a.
(a) 2 ≤ ab + ab , where ab > 0,
(b) ab + ab ≤ −2, where ab < 0.
a+b
2

≥

1.10. x1 ≤
1.11.

x1
y1

≤

1
a

x1 + ··· +xn
n
x1 + ··· +xn
y1 + ··· +yn

≤ xn , where x1 ≤ · · · ≤ xn .
≤

xn
,
yn

where

x1
y1

≤ ··· ≤

xn
yn

and yi > 0, i  1, . . . , n.

1
n

1.12. x1 ≤ (x1 · · · xn ) ≤ xn , where n ≥ 2, 0 ≤ x1 ≤ · · · ≤ xn .
1.13. |a1 | + · · · + |an | ≥ |a1 + a2 + · · · + an |.
1.14. a1 + ···n +an ≥ 1 + ···n + 1 , where n ≥ 2, ai > 0, i  1, . . . , n.
an
 a1
√
√
a+b
1.15. (a + b) 2 ≥ a b + b a, where a > 0, b > 0.

1.16. 21 (a + b) + 41 ≥ a +2 b , where a > 0, b > 0.
1.17. a(x + y − a) ≥ x y, where x ≤ a ≤ y.
1.18. x −1 1 + x +1 1 > 2x , where x > 1.
1.19.
1.20.

1
+ 3k1+ 2 + 3k1+ 3 > 2k1+ 1 + 2k1+ 2 , where
3k + 1
a)(1 − b)
ab
≤ ((1(1−−a)+(1
, where 0 < a ≤ 21 ,
− b))2
(a+b)2

k ∈ N.
0 < b ≤ 21 .

Problems

3

2k + 1
1.21. √3k1 + 1 · 2k
< √3k1 + 4 , where k ∈ N.
+2
1.22. 2n−1 ≥ n, where n ∈ N.
· 1 < 1.
1.23. 13 + 23 · 15 + 23 · 45 · 17 + · · · + 23 · 45 · 67 · · · 100
101 103
1−a
1−b
a
b
1.24. (a) 1 − b + 1 − a ≤ b + a , where 0 < a, b ≤ 21 ,
n
n
n
n


1 
1 
(b)
(1 − ai ) ≤
ai , where 0 < a1 , . . . , an ≤ 21 .
1 − ai
ai
i1

i1
i1
1
5
<
,
where
n
3
n
4
a+b
ab
+
,
where
2
3

1.25. 1 + 213 + · · · +
1.26. 1 + a1 + b ≤ 1 −

i1

∈ N.
0 ≤ a ≤ 1, 0 ≤ b ≤ 1.

Proofs
1.1. We have a 2 + b2 − 2ab  (a − b)2 ≥ 0. Note that equality holds if and only
if a  b.
√
1.2. The proof is similar to the proof of Problem 1.1. We have a + b − 2 ab 
√
√ 2
a − b ≥ 0. Note that equality holds if and only if a  b.
√

1.3. Multiplying both sides of the inequality of Problem 1.2 by 2a +ab
, we obtain the
b
required inequality. Note that equality holds if and only if a  b.
1.4. We have 2ab ≤ a 2 + b2 (see Problem 1.1). We obtain the equivalent inequality
2
2
2
a 2 + b2 + 2ab ≤ 2a 2 + 2b2 . Thus, it follows that (a +4b) ≤ a +2 b . The last


a+b
 ≤ a 2 + b2 . Therefore, a 2 + b2 ≥ a + b .
inequality can be rewritten as 
2

1.5.
1.6.
1.7.
1.8.

2

2

2

Note that equality holds if and only if a  b.
This inequality follows from the inequalities of Problems 1.2 and 1.3. Note
that equality holds if and only if a  b.
See the proof of Problem 1.4. Note that equality holds if and only if a  b.
Using that a + b − 1 − ab  (a − 1)(1 − b) and the condition b < 1 < a, it
follows that (a − 1)(1 − b) > 0.
Let us evaluate the difference between the left-hand side and the right-hand
side of the given inequality:
 


a 2 + b2 − c2 − (a + b − c)2  a 2 − c2 − (a + b − c)2 − b2 
 (a − c)(a + c) − (a − c)(a + 2b − c)  2(a − c)(c − b) > 0,

since according to the assumption, we have b < c < a.
1.9. (a) Multiplying both sides of the inequality by ab, where ab > 0, we deduce
that a 2 + b2 ≥ 2ab.
(b) Dividing both sides of the inequality a 2 +b2 ≥ −2ab by ab, where ab < 0,
we obtain the required inequality.
1.10. We have x1 ≤ x1 , . . . , xn ≤ xn , whence nx1 ≤ x1 + · · · + xn ≤ nxn .
Thus, it follows that x1 ≤ x1 + ···n +xn ≤ xn .

4

1 Basic Inequalities and Their Applications

1.11. From the assumptions of the problem, it follows that yi xy11 ≤ xi ≤ yi xynn , i 
1, . . . , n. Summing up these inequalities, we deduce that
x1
xn
(y1 + · · · + yn ) ≤ x1 + · · · + xn ≤ (y1 + · · · + yn ).
y1
yn
+xn
Hence, it follows that xy11 ≤ xy11 ++ ···
≤ xynn .
··· +yn
1.12. We have that x1 ≤ xi ≤ xn , i  1, . . . , n. Multiplying these inequalities, we
1
obtain x1n ≤ x1 · · · xn ≤ xnn . Thus, it follows that x1 ≤ (x1 · · · xn ) n ≤ xn .
1.13. If a1 + · · · + an ≥ 0, then |a1 + · · · + an |  a1 + · · · + an .
Using a ≤ |a|, we deduce that

|a1 + · · · + an |  a1 + · · · + an ≤ |a 1 | + · · · + |an |.
If a1 + · · · + an < 0, then |a1 + · · · + an |  −a1 − · · · − an . Using the inequality
−a ≤ |a|, it follows that |a1 + · · · + an |  −a1 − · · · − an ≤ |a 1 | + · · · + |an |.
1.14. We have that
⎛
⎞


⎜





1
an−1
1
an ⎟
⎜ a1 a2
⎟
⎜
+ ··· +
+ ··· +
+
+
(a1 + · · · + an )
⎟ + n,
⎝ a2 a1
a1
an
an
an−1 ⎠



n(n−1)/2

and using Problem 1.9 (a), it follows that



1
n(n − 1)
1
≥n+2·
 n2.
+ ··· +
(a1 + · · · + an )
a1
an
2


√
≥ 2 ab ·
√ √
√
a+ b
, which can be obtained by multiplying the inequalities a + b ≥ 2 ab
2

√
√
(Problem 1.2) and a +2 b ≥ a +2 b (see Problem 1.6).
1.16. We have
1.15. The given inequality is equivalent to the inequality (a + b)

1
1
(a + b) + −
2
4



a+b

2



a+b 1
−
2
2

a+b
2

2
≥ 0.


Therefore, 21 (a + b) + 41 ≥ a +2 b .
1.17. Since a(x + y − a) − x y  ax − x y + a(y − a)  (y − a)(a − x) and y ≥
a ≥ x, it follows that (y − a)(a − x) ≥ 0. Therefore, a(x + y − a) ≥ x y.
1.18. Using the inequality of Problem 1.5, it follows that

Proofs

5
1
x −1

+
2

1
x +1

>

2
1
2
1
+
> .
, or
x −1 x +1
x
(x − 1) + (x + 1)

1.19. According to the inequality of Problem 1.18, we have that
1
+ 3k1+ 3  (3k +12)−1 + (3k +12)+1 > 3k2+ 2 . Therefore, 3k1+ 1 +
3k + 1
3
.
3k + 2
Now let us prove that 3k3+ 2 > 2k1+ 1 + 2k1+ 2 .
Indeed, 2k1+ 1 + 2k1+ 2 − 3k3+ 2  (2k + 1)(2k−k+ 2)(3k + 2) < 0.

1
3k + 2

+

1
3k + 3

>

1.20. The given inequality is equivalent to the following inequality:


2
−1
a +b




2
≤





1
1
−1
−1 .
a
b

We have that


1
−1
a









2
1
2
1
1
4
1
4
−1 −
−1 
− − −

+
b
a +b
ab
a
b
(a + b)2 a + b

4
a +b
1
4
(a − b)2
(a − b)2 (1 − (a + b))
(a − b)2
−
−


+
−
2
2
ab
a +b
ab
ab(a + b)
ab(a + b)
ab(a + b)2
(a + b)

b) (1−(a + b))
and 0 < a ≤ 21 , 0 < b ≤ 21 . Then (a − ab(a
≥ 0, and therefore,
+ b)2
 2
2  1
 1

−1 ≤ a −1 b −1 .
a+b
√
1.21. The given
3k + 4 <
√ inequality is equivalent to the inequality (2k + 1)
(2k + 2) 3k + 1 or to the following inequality: (2k + 1)2 (3k + 4) <
(2k + 2)2 (3k + 1).
The last inequality holds because
2

(2k + 2)2 (3k + 1) − (2k + 1)2 (3k + 4)  k > 0.
1.22. Since 1 < 2 < 22 < · · · < 2n−1 and the number of positive integers
1, 2, 22 , . . . , 2n−1 is equal to n, it follows that 2n−1 ≥ n.
1.23. Consider a unit line segment and suppose on the first day, we paint 13 of thegiven
1
segment, the second day 15 of the rest of the segment, on the 51st day, 103
of the
rest of the segment. Since every day there remains a part of the given segment,
the sum of the painted parts must be less than 1.
The first day we have painted 13 of the given segment, on the second day 23 · 15 ,
on the 51st day 23 · 45 · · · 100
· 1 . Hence, we deduce that
101 103
1 2 1 2 4 1
2 4
98 100 1
+ · + · · + ··· + · ···
·
·
< 1.
3 3 5 3 5 7
3 5
99 101 103
1.24 (a) According to the inequality of Problem 1.20, we have that

6

1 Basic Inequalities and Their Applications

(a + b)2 1 − a 1 − b
a b
((1 − a) + (1 − b))2
≤
or
+
≤ + .
(1 − a)(1 − b)
ab
1−b 1−a
b a
An alternative proof. Note that 1 − a ≥ a and 1 − b ≥ b, and therefore,
1−a 1−b
(1 − a)2 + (1 − b)2
((1 − a) − (1 − b))2 + 2(1 − a)(1 − b)
+



1−b 1−a
(1 − a)(1 − b)
(1 − a)(1 − b)


(a − b)2
(a − b)2
a b
+2≤
+2 + .
(1 − a)(1 − b)
ab
b a

a
−b
Hence, we deduce that 11 −
+ 11 −
≤ ab + ab .
−b
a
n
n
 1 
(b) Since
(1 − ai ) 
1 − ai
i1
i1





1 − an−1
1 − a1 1 − a2
1 − an
+
+
+n, using the
+ ··· +
1 − a2 1 − a1
1 − an
1 − an−1



n(n−1)/2

inequality of Problem 1.24 (a), we obtain that






n
n
n


an−1
1
1 
a1 a2
an
+ ··· +
(1 − ai ) ≤
+
+
+n 
ai .
1 − ai
a2 a1
an
an−1
a
i1
i1
i1 i i1




n


n(n−1)/2

1.25. Note that if n ≥ 4, then
1+

1
2−1
n − (n − 1)
1
+ ··· + 3  1 +
+ ··· +
3
3
23
n
2

n



2
n−1
1
5
1
1
5
−
− ··· −
 −
−
−
< ,
3
2
3
2
3
4
2
3
3
4
n
4

since (k +k 1)3 > (k +12)2 , where k ∈ N.
1.26. We have that (1 − a)(1 − b) ≥ 0, and therefore a + b − 1 ≤ ab. Then



a+b
a+b
1
1
− 1−

(a + b − 1) ≤ ab.
1+a+b
2
2(1 + a + b)
3

Problems for Independent Study
Prove the following inequalities (1–32).
1. |x − y| < |1 − x y|, where |x| < 1, |y| < 1.
sin x−1
x
2. sin
+ 21 ≥ 2−sin
.
x−2
3−sin x
a
b
c
2
3. bc + ca + ab ≥ a + b2 − 2c , where a > 0, b > 0, c > 0.
1
4. a1 + b1 − 1c < abc
, where a 2 + b2 + c2  53 and a > 0, b > 0, c > 0.



2
5. 3 1 + a 2 + a 4 ≥ 1 + a + a 2 .

Problems for Independent Study

7

6. (ac + bd)2 + (ad − bc)2 ≥ 144, where a + b  4 , c + d  6 .
√
2
+ na 2√≥ a 2(x1 + x2 + · · · + x2n ).
7. x12 + x22 + · · · + x2n
8.

1
+ b +1 c + a 1+ c ≤
a +b


3 2
2
3 2

√
√
a+ b+ c
√
, where a >
2 
abc


2
3 2
2

0, b > 0, c > 0.

9. a b − c + b c − a + c a − b < 0, where 0 < a < b < c.
10. a 3 b + b3 c + c3 a ≥ a 2 b2 + b2 c2 + a 2 c2 , where a ≥ b ≥ c > 0.
2
11. xy + yz + x +y z ≤ (x x+zz) , where 0 < x ≤ y ≤ z.





√
√
√
12.
1 + a + 1 + a + a 2 + · · · + 1 + a + · · · + a n < na,
where n ≥ 2, a ≥ 2, n ∈ N.
1)2
, where tan α  n tan β , n > 0 ,
13. (a) tan2 (α − β) ≤ (n −
4n
(b) 1 + cos(α − β) ≥ cos α + cos β, where 0 ≤ α ≤ π2 , 0 ≤ β ≤ π2 .
+ [3x]
+ [4x]
+ [5x]
,
14. [5x] ≥ [x] + [2x]
2
3
4
5
where [a] is the integer part of the real number a.
15. (n!)2 ≥ n n , where n ∈ N.
16. x 6 + x 5 + 4x 4 − 12x 3 + 4x 2 + x + 1 ≥ 0.
17. log2 α ≥ log β log γ , where α > 1, β > 1, γ > 1, α 2 ≥ βγ .
18. log4 5 + log5 6 + log6 7 + log7 8 > 4, 4.
n
< 21 , where n ∈ N.
19. 13 + · · · + 3·5···(2n
+ 1)
3

3

20. 223 −+ 11 · · · nn3 −+ 11 < 23 , where n ≥ 2, n ∈ N.
21. 1 · 1! +
· · · + n ·n! < (n + 1)!
 , where n ∈ N.
22. 1 + 212 1 + 312 . . . 1 + n12 < 2, where n ≥ 2, n ∈ N.


 

23. 1 − p12 1 − p12 . . . 1 − p12 > 21 , where 1 < p1 < p2 < · · · < pn , pi ∈
1

n

2

N , i  1, . . . , n.
1
1
+ 1000
< 25 .
24. 21 − 13 + 41 − 15 + · · · − 999
25. (sin x + 2 cos 2x)(2 sin 2x − cos x) < 4, 5.
26. (a) 1 +a a+ +b b ≤ 1 +a a + 1 +b b , where a ≥ 0, b ≥ 0.


(b) 2 +a a+ +b b ≥ 21 1 +a a + 1 +b b , where a ≥ 0, b ≥ 0.
(c) n ≤

a1 + b1
a1 + b1 + 2

+ ... +

an + bn
an + bn + 2

+

1
a1

1
a1

+

1
bi
1
1
bi + 2
1

+

+ ... +

1
1
an + bi n
1
1
+
an
bi n + 2

< 2n, where

i 1 , . . . , i n is some permutation of the numbers 1, . . . , n, ai , bi > 0, i 
1, . . . , n.
n
n


a1 + 2a2 + ··· + iai
≤
2
ai , where ai ≥ 0, i  1, . . . , n.
27.
2
i
i1

i1

41
, where a1 + b1 + 1c < 1 , a, b, c ∈ N .
28. a1 + b1 + 1c ≤ 42
y
4x
z
29. y + z + x + z + x + y > 2, where x, y, z > 0.
30. 1 < a + ab + d + a + bb + c + b + cc + d + a + dc + d < 2, where a > 0, b > 0, c > 0, d > 0.
31. a + b > c + d, where a, b, c, d ≥ 21 and a 2 + b > c2 + d, a + b2 > c + d 2 .
Hint. If a+b ≤ c+d, then a ≤ c or b ≤ d. If a ≤ c, then b−d > (c−a)(c+a) ≥
c − a.
√
32. (b − a)(9 − a 2 ) + (c − a)(9 − b2 ) + (c − b)(9 − c2 ) ≤ 24 2, where 0 ≤ a ≤
b ≤ c ≤ 3.

8

1 Basic Inequalities and Their Applications

Hint. We have that
(b − a)(9 − a 2 ) + (c − a)(9 − b2 ) + (c − b)(9 − c2 ) ≤ 9b + c(9 − b2 ) + (c − b)(9 − c2 ) 
 18c − c3 + bc(c − b) ≤ 18c − c3 +

1 3
3
c  18c − c3 .
4
4

33. If 0 < a, b, c < 1, then one of the numbers (1 − a)b, (1 − b)c, (1 − c)a is
not greater than 41 .
34. Let a > 0, b > 0, c > 0, and a + b + c  1. Prove that



(a)
a + 41 (b − c)2 + b + 41 (c − a)2 + c + 41 (b − a)2 ≤ 2,

√
√
√
(b)
a + 41 (b − c)2 + b + c ≤ 3.

Hint. (a) x + 41 (y − z)2 ≤ x + y+z
if x, y, z > 0 and x + y + z  1.
2
35. Find the smallest possible value of the following expression:
4
2
2
a4
+ ab4 − ab2 − ab2 + ab + ab , where a > 0, b > 0.
b4

Chapter 2

Sturm's Method

Historical origins. Sturm's method was proposed by the Prussian mathematician
Friedrich Otto Rudolf Sturm, born 6 January 1841 in Breslau, Prussia (now Wrocław,
Poland), died 12 April 1919 in Breslau, Germany. Sturm obtained his doctorate
in mathematics from the University of Breslau (now University of Wrocław) in
1863 under the supervision of the well-known Prussian mathematician Heinrich
Eduard Schröter. Sturm was the advisor of 19 doctoral students, including the wellknown Prussian mathematician Otto Toeplitz, born 1 August 1881 in Breslau, Prussia,
died 15 February 1940 in Jerusalem, Mandatory Palestine (now Israel). Toeplitz
studied mathematics at the University of Breslau, where under Sturm's supervision
he obtained his doctorate in 1905.
Besides its various applications, Sturm's method provides an opportunity to prove
a large number of different inequalities under certain conditions.
Example 2.1 Prove that if the product of positive numbers x1 , . . . , xn (n ≥ 2) is
equal to 1, then x1 + · · · + xn ≥ n.
Proof If x1  · · ·  xn  1, then x1 + · · · + xn  n.
Suppose that among the considered numbers there are at least two different numbers. Then among the numbers x1 , . . . , xn there are two numbers such that one of
them is greater than 1 and the other one is less than 1 (Problem 1.12). Without loss of
generality one can assume that those numbers are x1 and x2 , and that x1 < 1 < x2 .
Note that x1 + x2 > 1 + x1 x2 (Problem 1.7).
If one substitutes the given numbers by numbers 1, x1 x2 , x3 , . . . , xn , then their
product is again equal to 1, and the sum satisfies 1+ x1 x2 + x3 +· · ·+ xn < x1 +· · ·+ xn .
Doing the same with the numbers 1, x1 x2 , x3 , . . . , xn , in a similar way we obtain
a new sequence such that two numbers in it are equal to 1. Doing the same at most
n − 1 times, we obtain a sequence such that n − 1 numbers in it are equal to 1, and
the nth number is equal to x1 · · · xn .
On the other hand, x1 · · · xn  1.

© Springer International Publishing AG, part of Springer Nature 2018
H. Sedrakyan and N. Sedrakyan, Algebraic Inequalities, Problem Books
in Mathematics, https://doi.org/10.1007/978-3-319-77836-5_2

9

10

2 Sturm's Method

Hence, we obtain that n < x1 + x2 + · · · + xn .
From the proof, it follows that equality holds if and only if x1  · · ·  xn  1.
Example 2.2 Prove that if the sum of the numbers x1 , . . . , xn (n ≥ 2) is equal to 1,
then x12 + · · · + xn2 ≥ n1 .
Proof If x1  · · ·  xn  n1 , then x12 + · · · + xn2  n1 .
Suppose that among the considered numbers there are at least two different numbers.
Then among the numbers x1 , . . . , xn there are two numbers such that one of them
is greater than n1 and the other one is less than n1 (Problem 1.10). Without loss of
generality, one can assume that those numbers are x1 and x2 , and that x1 < n1 and
x2 > n1 . Therefore, substituting x1 by n1 , and x2 by x1 + x2 − n1 , we obtain a new
sequence of numbers n1 , x1 + x2 − n1 , x3 , . . . , xn such that their sum is again equal to 1.
On the other
2  we have that
2
 hand,
x12 + x22 > n1 + x1 + x2 − n1 (see Problem 1.8), whence
x12

+ ··· +

xn2


 2 
1
1 2
>
+ x1 + x2 −
+ x32 + · · · + xn2 .
n
n

Repeating these steps a finite number of times, we obtain a sequence such that
its all terms are equal to n1 , and the sum of their squares is less than the sum of the
 2
 2
squares of the numbers x1 , . . . , xn , that is, x12 + · · · + xn2 > n1 + · · · + n1  n1 .
From the proof, it follows that equality holds if and only if x1  · · ·  xn  n1 .

Problems
Prove the following inequalities (2.1–2.6).
2.1 AM-GM (arithmetic mean–geometric mean inequality):
√
x1 + ··· + xn
≥ n x1 · · · xn , where n ≥ 2, xi > 0, i  1, . . . , n.
n
2.2 QM-AM (quadratic
mean–geometric mean inequality or RMS/root mean square

2.3
2.4
2.5

x12 + ··· + xn2
n

≥ x1 + ···n + xn .
(1 − x1 ) ··· (1 − xn )
≥ (n − 1)n , where n ≥ 2, xi > 0, i  1, . . . , n, and
x1 ··· xn
x1 + · · · + xn  1.
1
1
+ · · · + 1+x
≥ 1+ √n xn1 ... xn , where n ≥ 2, x1 ≥ 1, . . . , xn ≥ 1.
1+x1
n
1
abc + bcd + cda + dab ≤ 27
+ 176
abcd, where a ≥ 0, b ≥ 0, c ≥ 0, d ≥ 0,
27
inequality):

and a + b + c + d  1.
2.6 0 ≤ x y + yz + zx − 2x yz ≤

7
,
27

where x ≥ 0, y ≥ 0, z ≥ 0 and x + y + z  1.

Problems

11

2.7 Among all triangles with no angle greater than 75º inscribed in a given circle,
find the triangle such that its perimeter is
(a) the greatest,
(b) the smallest.
2.8 (a) Schur's inequality: Prove that if for some numbers α and β one has the
inequality [α f (a) + β f (b) ≤ f (αa + βb)]α f (a) + β f (b) ≥ f (αa + βb),
where α ≥ 0, β ≥ 0, α + β  1, a, b are any numbers belonging to
D( f )  I,1 and x1 , . . . , xn , y1 , . . . , yn ∈ I such that y1 ≥ . . . ≥ yn ,y1 ≤
x1 , y1 + y2 ≤ x1 + x2 , . . . , y1 + · · · + yn−1 ≤ x1 + · · · + xn−1 , y1 + · · · + yn 
x1 + · · · + xn , then f (y1 ) + f (y2 ) + · · · + f (yn ) ≤ f (x1 )+ f (x2 ) + · · · + f (xn )
[ f (y1 ) + f (y2 ) + · · · + f (yn ) ≥ f (x1 ) + f (x2 ) + · · · + f (xn )].
(b) Popoviciu's inequality: Prove that if for numbers α and β one has the
inequality α f (a) + β f (b) ≥ f (αa + βb), where α, β ≥ 0, α + β  1,
and a, b are any numbers belonging to the interval I , then for all numbers
x, y, z from the interval I , one has the inequality
f (x) + f (y) + f (z) + 3 f

x + y + z
x + y
y +z
z + x 
≥2f
+2f
+2f
.
3
2
2
2

2.9 Suppose that for numbers x1 , . . . , x1997 , the following conditions hold:
√
(a) − √13 ≤ xi ≤ 3, i  1, . . . , 1997,
√
(b) x1 + · · · + x1997  −318 3.
12
.
Find the greatest possible value of the expression x112 + · · · + x1997

n−1
2
...