Sample problems from Olympiad Inequalities Book ?· Sample problems from Olympiad Inequalities Book…
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Sample problems fromOlympiad Inequalities Book
This book is intended as a useful resource for high school and college stu
dents who are training for national or international mathematical competi
tions. But anybody who is interested in elementary mathematical inequali
ties may find this book useful. This problem solving book is divided into
six chapters, containing more than fifty topics of interest to mathematical
olympiad contestants and coaches, demonstrating ideas and strategies in solv
ing inequalities. The reader will find in the book clever applications of well
known results as well as powerful original methods, each is explained and
illustrated by carefully selected problems.
Chapter 1 Inequalities between means
This chapter starts with the fundamental fact x2 0, upon which many interestinginequalities are derived. It also serves to emphasize that powerful results can be obtained
by little means.
Chapter 2 CauchySchwars
The classical result of CauchySchwarz is revisited with examples illustrating sophis
ticated, at time surprising, ways to apply the inequality. Other classical inequalities such
as those of Chebyshev and Holder are also discussed and shown how they might cooperate
with CauchySchwarz inequality.
Chapter 3 Convexity
This chapter utilises calculus in solving inequalities. Based on simpe properties of
linear and convex functions, systematic methods are derived to tackle some advanced
problems. Also discussed is tangent line method, which gives a geometric interpretation
of bounds.
1
Chapter 4 Homogenous inequalities
Homogeneous inequalities constitutes a large class of inequality problems. This chap
ter discusses various approaches to solving this class of inequalities, including the tech
niques of homogenization, normalisation, the application of Rolles theorem to reduce
the number of variables, the use of limits and partitions, quadratic estimations, and estab
lishing new bounds through isolated fudging. Especially in focus are powerful techniques
to solve inequalities by the change of variables p = a + b + c, q = ab + bc + ca andr = abc and by transforming them to one of the following forms
(1) x(a b)(a c) + y(b c)(b a) + z(c a)(c b) 0,
(2) x(a b)2 + y(b c)2 + z(c a)2 0,
(3) M(a b)2 + N(c a)(c b) 0.
All three, four variable symmetric polynomial inequalities can be solved using ideas in
this chapter.
Chapter 5 The method of Mixing Variables
The method of mixing variables has been used in various forms for decades  an ex
ample is G. Polyas delightful proof of the AMGM inequalities. This chapter examines
this idea in depth with extension in different directions. The first three sections explain
why mixing variables work, give hints to find approriate variables to mix by taking equal
ity cases into consideration. The most important results in this chapter are two theorems
which facilitates solutions for a large class of multivariable inequalities.
Chapter 6 Further Topics and problems with solutions
The chapter starts with miscellaneous indenpendent topics touching upon various
aspects of solving inequalities. The discussion includes the interplay between trogono
metric and algebraic substitution, absolute values, inequalities with special equality cases
and inequalities with ordered sequences.
Authors: Phm Vn Thun, L V
Hanoi University of Science, Vietnam
332 pages, LATEX typset, soft cover
Price: 12 (twelve) USD
It is available for sale in La Thanh Hotel where deputy leaders and contestants stay.
3
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b
b
Phm Vn Thun, L V
Olympiad Inequalities
Introduction to the art of solving inequalities
bbb
f (x) = (x a)(x b)(x c)
y = f (x)
b
b
O
a b c x
y
f
(
1 t3
)
f
(
1 + t
3
)
VIETNAM NATIONAL UNIVERSITY PRESS
5
The 11 out of 600 problems
Problem 1. Prove that if x, y, z are real numbers, then
7(x4 + y4 + z4) + 10(x3y + y3z + z3x) 0.
Problem 2. Prove that a, b, c are positve real numbers, then
a
b2 + 14 bc + c2
+b
c2 + 14 ca + a2
+c
a2 + 14 ab + b2 2.
Problem 3. Let a, b, c, d be nonnegative real numbers such that
a2 + b2 + c2 + d2 = 1.
Prove that
a + b + c + d a3 + b3 + c3 + d3 + ab + bc + cd + da + ac + bd.
Problem 4. Suppose that p, q, r, s are real numbers such that the following equation hasfour roots (not neccessarily distinct)
x4 px3 + qx2 rx + s = 0.
Prove that (p2 2q)5/2 + 8ps 4(p2 2q)r.
Problem 5. Let n be a positive integers, n 2. Nonnegative real numbers a1, a2, ..., ansatisfy a1 + a2 + + an = s, s < 2, define
f (a1, a2, ..., an) = 1i< jn
1
1 (
ai+a j2
)2.
Prove that
1
2n(n 1)/
[
1 (
s
n
)2]
f (a1, a2, ..., an) n 1
1 s2/4+
1
2(n 1)(n 2).
Determine cases of equality.
Problem 6. Let a, b, c, d be nonnegative real numbers such that a + b + c + d = 2.Prove that
ab(a2 + b2 + c2) + bc(b2 + c2 + d2) + cd(c2 + d2 + a2) + da(d2 + a2 + b2) 2.
Problem 7. Prove that if x, y, z are postive real numbers then
x
y+
y
z+
z
x
(
x2 + y2 + z2
xy + yz + zx
)2/3
.
6
Problem 8. Let r, a, b, c be positive real numbers, put p = 2r 3
r + 2. Prove that
a
pa + rb + c+
b
pb + rc + a+
c
pc + ra + b
1
1
r + r.
Problem 9. Prove that if a, b, c, d are nonnegative real numbers then
1
a2 + b2 + c2+
1
b2 + c2 + d2+
1
c2 + d2 + a2+
1
d2 + a2 + b2
12
(a + b + c + d)2.
Problem 10. Let x, y, z be nonnegative real numbers such that x2 + y2 + z2 = 1. Provethat
cyclic
1 xy.
1 yz 2.
Problem 11. Prove that if x, y, z R then
x(x + y)3 + y(y + z)3 + z(z + x)3 8
27(x + y + z)4.
Do you think problem 10 can be solved using only CauchySchwarz inequal
ity? If not, have a look at this book. Do you believe that a solution for problem 9
is just a few line long with only simple reasoning? Problem 4 and 5 look intimi
dating but we have strategies to deal with such types. Problem 3 is selected from
the section on symmetric polynomials which contains a large number of original
and nice inequalities. There is a very short and clever solution to problem 6. Can
you figure it out? Problem 11 is a refinement of a wellknown inequality.
This book offers a lot more beautiful problems together with powerful meth
ods and strategies.

Do not miss this book when you are in Hanoi for IMO.
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