Straight from the boook

Straight
from
the
Book
Titu
Andreescu
Gabriel
Dospinescu
Straight
from
the
Book
XYZ~
Press
Titu
Andreescu
Gabriel Dospinescu
University of Texas
at
Dallas
Ecole Normale Superieure, Lyon
Library
of
Congress
Control
Number:
2012951362
ISBN-13: 978-0-9799269-3-8 ISBN-IO: 0-9799269-3-9
© 2012 XYZ Press, LLC
All rights reserved.
This

work
may
not
be
translated
or copied in whole or in
part
without
the
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publisher (XYZ Press, LLC, 3425
Neiman Rd.,
Plano,
TX
75025, USA)
and
the
authors
except for brief excerpts
in connection
with
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with
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The
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trademarks,
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to
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9 8 7 6 5 4 3 2 1
www.awesomemath.org
Cover design by
Iury
Ulzutuev
The only way to learn mathematics is to
do
mathematics.
,
-Paul
Halmos
Foreword
This
book
is a follow-on from
the
authors'
earlier book,
'Problems
from
the
Book'. However,

it
can
certainly
be
read
as a
stand-alone
book:
it
is
not
vital
to
have
read
the
earlier book.
The
previous
book
was
based
around
a collection
of
problems.
In
contrast,
this
book

is
based
around
a collection
of
solutions.
These
are
solutions
to
some of
the
(often extremely challenging) problems from
the
earlier book.
The
topics chosen reflect those from
the
first twelve
chapters
of
the
previous
book: so
we
have Cauchy-Schwarz, Algebraic
Number
Theory,
Formal
Series,

Lagrange
Interpolation,
to
name
but
a few.
The
book
is one
of
the
most
remarkable
mathematical
texts
I have ever
seen.
First
of all,
there
is
the
richness
of
the
problems,
and
the
huge variety
of solutions.

The
authors
try
to
give several solutions
to
each problem,
and
moreover give insight
about
why each
proof
is
the
way
it
is,
in
what
way
the
solutions differ from each
other,
and
so on.
The
amount
of
work
that

has
been
put
in,
to
compile
and
interrelate
these solutions, is simply staggering.
There
is enough here
to
keep
any
devotee of problems going for years
and
years.
Secondly,
the
book
is far more
than
a collection of solutions.
The
solu-
tions are used as
motivation
for
the
introduction

of some very clear expositions
of
mathematics.
And
this
is modern, current,
up-to-the-minute
mathematics.
For example, a discussion of
Extremal
Graph
Theory
leads
to
the
celebrated
Szemen§di-
Trotter
theorem
on
crossing numbers,
and
to
the
amazing applica-
tions
of
this
by
Szekely

and
then
on
to
the
very recent
sum-product
estimates
of Elekes, Bourgain,
Katz,
Tao
and
others.
This
is absolutely
state-of-the-art
material.
It
is presented very clearly:
in
fact,
it
is
probably
the
best
exposition
of
this
that

I have seen
in
print.
viii
As
another
example,
the
Cauchy-Schwarz section leads on
to
the
devel-
opements
of sieve theory, like
the
Large Sieve of Linnik
and
the
Tunin-Kubilius
inequality.
And
again, everything is incredibly clearly presented.
The
same
applies
to
the
very large sections on Algebraic
Number
Theory,

on
p-adic
Analysis,
and
many
others.
It
is
quite
remarkable
that
the
authors
even know so much
current
math-
ematics
I do
not
think
any
of my colleagues would
be
so well-informed over
so wide
an
area.
It
is also remarkable
that,

at
least in
the
areas in which I
am
competent
to
judge,
their
explanations of these topics are polished
and
exceptionally well
thought-out:
they
give
just
the
right words
to
help someone
understand
what
is going on.
Overall,
this
seems
to
me like
an
'instant

classic'.
There
is so much
material, of such a high quality, wherever one
turns.
Indeed, if one opens
the
book
at
random
(as I have done several times), one is pulled
in
immediately
by
the
lovely exposition. Everyone who loves
mathematics
and
mathematical
thinking
should acquire
this
book.
Imre
Leader
\
Professor of
Pure
Mathematics
University of Cambridge

Preface
This
book
is a compilation
of
many
suggestions, much advice,
an
even
more
hard
work.
Its
main
objective is
to
provide solutions
to
the
problems
which were originally proposed in
the
first
12
chapters
of Problems from the
Book.
We were
not
able

to
provide full solutions
in
our
first volume
due
to
the
lack of space.
In
addition,
the
statements
of
the
proposed
problems
contained'
typos
and
some
elementary
mistakes which needed
further
editing. Finally,
the
problems were also considered
to
be
quite

difficult
to
tackle.
With
these
points in mind,
we
came
up
with
a
two-part
plan:
to
correct
the
identified
errors
and
to
publish comprehensive solutions
to
the
problems.
The
first task, editing
the
statements
of
the

proposed problems, was sim-
ple
and
has already
been
completed
in
the
second edition
of
Problems from the
Book.
Although
we focused
on
changing several problems,
we
also
introduced
many
new ones.
The
second
task,
providing full solutions, however, proved
to
be
more challenging
than
expected, so we asked for help. We

created
a forum
on
www.mathlinks.ro
(a
familiar site for problem-solving enthusiasts) where
solutions
to
the
proposed problems were
gathered.
It
was a
great
pleasure
to
witness
the
passion
with
which some of
the
best
problem-solvers
on
mathlinks
attacked
these tough-nuts.
This
new

book
is
the
result
of
their
common
effort,
and
we
thank
them.
Providing solutions
to
every problem
within
the
limited space of one vol-
ume
turned
out
to
be
an
optimistic plan.
Only
the
solutions
to
problems

from
the
first 12
chapters
of
the
second edition of Problems from the Book
are
presented here.
Furthermore,
many
of
the
problems
are
difficult
and
require
a
rather
extensive
mathematical
background. We decided, therefore,
to
com-
plement
the
problems
and
solutions

with
a series of
addenda,
using various
problems as
starting
points for excursions
into
"real
mathematics".
Although
x
we never
underestimated
the
role
of
problem solving, we strongly believe
that
the
reader
will benefit more from embarking on a
mathematical
journey
rather
than
navigating a huge list of
scattered
problems.
This

book
tries
to
reconcile
problem-solving
with
"professional
mathematics"
.
Let
us now delve
into
the
structure
of
the
book
which consists of
12
chap-
ters
and
presents
the
problems
and
solutions proposed
in
the
corresponding

chapters
of
Problems from the Book, second edition.
Many
of
these
problems
are fairly difficult
and
different approaches are presented for a
majority
of
them.
At
the
end
of each
chapter,
we
acknowledge
those
who provided so-
lutions. Some
chapters
are followed
by
one
or
two
addenda,

which present
topics of more advanced
mathematics
stemming
from
the
elementary topics
discussed
in
the
problems of
that
chapter.
The
first two
chapters
focus
on
elementary algebraic inequalities
(a
no-
table
caveat is
that
some of
the
problems are
quite
challenging)
and

there
is
not
much
to
comment
regarding these
chapters
except for
the
fact
that
al-
gebraic inequalities have proven fairly
popular
at
mathematics
competitions.
To provide relief from
this
rather
dry
landscape
(the
reader
will notice
that
most
of
the

problems in
these
two
chapters
start
with
"Let
a,
b,
c
be
positive
real
numbers"),
we
included
an
addendum
presenting deep applications of
the
Cauchy-Schwarz inequality
in
analytic
number
theory. For instance,
we
dis-
cuss Gallagher's sieve, Linnik's large sieve
and
its version

due
to
Montgomery.
We
apply
these results
to
the
distribution
of prime numbers (for instance
Brun's
famous
theorem
stating
that
the
sum
of
the
inverses of
the
twin
primes
converges).
The
note-worthy Tunin-Kubilius inequality
and
its
classical appli-
cations

(the
Hardy-Ramanujan
theorem
on
the
distribution
of prime factors of
n,
Erdos'
multiplication
table
problem, Wirsing's generalization of
the
prime
number
theorem)
are
also discussed.
The
reader will
be
exposed
to
the
power
of
the
Cauchy-Schwarz inequality
in
real

mathematics
and,
hopefully, will un-
derstand
that
the
gymnastics
of
three-variables algebraic inequalities is
not
the
Holy Grail.
Chapter
3 discusses problems
related
to
the
unique factorization
of
inte-
gers
and
the
p-adic valuation maps. Among
the
topics discussed,
we
cover
the
local-global principle (which is extremely helpful

in
proving divisibilities
or
arithmetic
identities), Legendre's formula giving
the
p-adic valuation of n!,
xi
a beneficial
elementary
result called lifting the exponent lemma, as well as
more advanced techniques from p-adic analysis.
One
of
the
most
beautiful re-
sults
presented
in
this
chapter
in
the
celebrated Skolem-Mahler-Lech
theorem
concerning
the
zeros of a linearly recursive sequence. Readers will
perhaps

appreciate
the
applications
of
p-adic analysis, which is covered extensively in
a long
addendum.
This
addendum
discusses, from a foundational level,
the
arithmetic
of
p-adic numbers, a
subject
that
plays a
central
role
in
modern
number
theory. Once
the
basic groundwork
has
been
laid, we discuss
the
p-adic analogues of classical functions (exponential, logarithm,

Gamma
func-
tion)
and
their
applications
to
difficult congruences (for instance, Kazandzidis'
famous supercongruence).
This
serves as a good
opportunity
to
explore
the
arithmetic
of
Bernoulli numbers, Volkenborn's
theory
of
p-adic integration,
Mahler
and
Amice's classical theorems characterizing continuous
and
locally
analytic functions
on
the
ring

of
p-adic integers
or
Morita's
construction
of
the
p-adic
Gamma
function. A second
addendum
to
this
chapter
discusses various
classical
estimates
on
prime numbers, which
are
used
throughout
the
book.
Chapter
4 discusses problems
and
elementary
topics
related

to
prime num-
bers
of
the
form
4k
+ 1
and
4k
+
3.
The
most
intriguing problem discussed
is,
without
any
doubt,
Cohn's
renowned
theorem
characterizing
the
perfect
squares in
the
Lucas sequence.
Chapter
5 is

dedicated
again
to
the
yoga of
algebraic inequalities
and
is followed
by
an
addendum
discussing applications
of Holder's inequality.
Chapter
6 focuses
on
extremal
graph
theory. Most
of
the
problems revolve
around
Turan's
theorem, however
the
reader
will also
be
exposed

to
topics
such as chromatic numbers,
bipartite
graphs, etc.
This
chapter
is followed
by
a relatively advanced
addendum,
discussing various topics
related
to
the
Szemeredi-
Trotter
theorem, which gives
bounds
for
the
number
of
incidences
between a set of points
and
a
set
of curves. We discuss
the

theorem's
classical
probabilistic proof,
its
generalization
to
multi-graphs
due
to
Szekely
and
its
application
to
the
sum-product
problem
due
to
Elekes, as well as more recent
developments
due
to
Bourgain,
Katz,
and
Tao.
These
results are
then

applied
to
natural
and
nontrivial geometric questions (for instance:
what
is
the
least
number
of
distinct
distances
determined
by
n
points
in
the
plane?
what
is
the
maximal
number
of triangles
of
the
same
area?). Finally,

another
addendum
Xll
completes
this
chapter
and
is
dedicated
to
the
powerful probabilistic
method.
After
a
short
discussion
on
finite probability spaces, we provide
many
examples
of combinatorial applications.
Chapter
7 involves combinatorial
and
number
theoretic applications
of
finite Fourier analysis.
The

central
principles of
this
chapter
include
the
roots
of
unity
and
the
fact
that
congruences between integers
can
be
expressed in
terms
of
sums
of powers of
roots
of unity. To provide
the
reader
with
a
broader
view,
we

beefed-up
this
chapter
with
an
addendum
discussing Fourier analysis
on
finite abelian groups
and
applying
it
to
Gauss
sums
of Dirichlet characters,
additive problems, combinatorial or analytic
number
theory. For instance,
the
reader
will find a discussion of
the
P6lya-Vinogradov inequality
and
Vina-
gradov's beautiful use of
this
inequality
to

deduce a
rather
strong
bound
on
the
least
quadratic
non-residue
mod
p.
At
the
same
time, we explore Dirich-
let's
L-functions,
culminating
in
a
proof
of Dirichlet's
theorem
on
arithmetic
progressions.
This
section is
structured
to

first present
the
usual analytic
proof
(since
it
is really a masterpiece from all points of view),
up
to
some
important
simplification
due
to
Monsky.
At
the
same
time,
we
also discuss how
to
turn
this
into
an
elementary
proof
that
avoids complex analysis

and
dramatically
uses Abel's
summation
formula.
Chapter
8 focuses
on
diverse applications
of
generating functions.
This
is
an
absolutely crucial tool
in
combinatorics,
be
it
additive or enumerative.
In
this
section,
the
reader
will have
the
opportunity
to
explore enumerative

problems
(Catalan's
problem, counting
the
number
of solutions of linear dio-
phantine
equations
or
the
number
of irreducible polynomials
mod
p, of fixed
degree). We also discuss exotic combinatorial identities or recursive sequences,
which
can
be
solved elegantly using generating functions,
but
also
rather
chal-
lenging congruences
that
appear
so often
in
number
theory.

The
chapter
is
followed
by
an
addendum
presenting a very classical topic
in
enumerative
combinatorics, Lagrange's inversion formula.
Among
the
applications, let us
mention Abel's identity,
other
derived combinatorial identities, Cayley's
the-
orem
on
labeled trees,
and
various
related
problems.
Chapter
9 is
rather
extensive,
due

to
the
vast
nature
of
the
topic covered,
algebraic
number
theory. While
we
are able only
to
scratch
the
surface, nev-
ertheless
the
reader
will find a variety of intriguing techniques
and
ideas. For
xiii
instance, we discuss
arithmetic
properties
of
cyclotomic polynomials (includ-
ing
Mann's

beautiful
theorem
on
linear
equations
in
roots
of
unity),
rationality
problems,
and
various applications of
the
theorem
of
symmetric
polynomials.
In
addition,
we
present techniques
rooted
in
the
theory
of ideals
in
number
fields, finite fields

and
p-adic
methods.
We also give
an
overview of
the
el-
ementary
algebraic
number
theory
in
the
addendum
following
this
chapter.
After
a brief review
on
ideals, field extensions,
and
algebraic numbers,
we
proceed
with
a discussion of
the
primitive element

theorem
and
embeddings
of
number
fields
into
C.
We also briefly survey Galois theory,
and
the
fun-
damental
theorem
on
the
prime
factorization
of
ideals
in
number
fields,
due
to
Kummer
and
Dedekind. Once
the
foundation has

been
set, we discuss
the
prime factorization
in
quadratic
and
cyclotomic fields
and
apply
these tech-
niques
to
basic problems
that
explore
the
aforementioned theories. Finally,
we
discuss various applications of
Bauer's
theorem
and
of
Chebotarev's
theorem.
The
next
addendum
is concerned

with
the
fascinating topic of counting
the
number
of solutions modulo p of systems of polynomial equations. We use
this
as
an
opportunity
to
state
and
prove
the
basic
structural
results on finite
fields
and
introduce
the
Gauss
and
Jacobi
sums. We go
ahead
and
count
the

number
of
points
over a finite field of a diagonal hypersurface
and
to
compute
its
zeta
function.
This
is a beautiful
theorem
of Weil,
the
very
tip
of a massive
iceberg.
Chapter
10 focuses on
the
arithmetic
of polynomials
with
integer coeffi-
cients.
An
essential aspect of
the

discussion
concentrates
on
Mahler expan-
sions,
the
theory
of finite differences,
and
their
applications.
The
techniques
used in this
chapter
are
rather
diverse.
Although
the
problems
can
be
consid-
ered basic,
they
are challenging
and
require advanced problem-solving skills.
Chapter

11
provides respite from
the
difficult
tasks
mentioned
above.
It
discusses Lagrange's
interpolation
formula, allowing a unified
presentation
of
various
estimates
on
polynomials.
The
longest
and
certainly
most
challenging
chapter
is
the
last
one.
It
explores several algebraic techniques in combinatorics.

The
methods
are
stan-
dard
but
powerful.
The
last
part
of
the
chapter
deals
with
applications
to
geometry, presenting some of
Dehn's
wonderful ideas.
The
last problem pre-
sented
in
the
book
is
the
famous Freiling, Laczkovich, Rinne, Szekeres theorem,
xiv

a
stunningly
beautiful application of algebraic combinatorics.
We would like
to
thank,
again,
the
members
of
the
mathlinks
site
for
their
invaluable
contribution
in
providing solutions
to
many
of
the
problems
in
this
book. Special
thanks
are
due

to
Richard Stong, who
did
a remarkable
job
by
pointing
out
many
inaccuracies
and
suggesting numerous
alternative
solutions. We would also like
to
thank
Joshua
Nichols-Barrer,
Kathy
Cordeiro
and
Radu
Sorici, who gave
the
manuscript
a readable form
and
corrected
several infelicities.
Many

of
the
problems
and
results
in
this
book
were used by
the
authors
in courses
at
the
AwesomeMath
Summer
Program,
and
students'
reactions guided us
in
the
process of simplifying
or
adding
more details
to
the
discussed problems. We wish
to

thank
them
all, for
their
courage
in
taking
and
sticking
with
these
courses, as well as for
their
valuable suggestions.
Titu
Andreescu

Gabriel Dospinescu

Contents
1
Some
Useful
Substitutions
1.1
The
relation a
2
+ b
2

+ e
2
= abc + 4 . . . . . . . . . . . . . . .
1.2
The
relations abc = a + b + e + 2
and
ab + be +
ea
+ 2abe = 1
1.3
The
relation a
2
+ b
2
+ e
2
+ 2abe = 1
1.4 Notes

.
1
2
9
21
27
2
Always
Cauchy-Schwarz.

. .
29
2.1 Notes

61
Addendum
2.A Cauchy-Schwarz
in
Number
Theory
62
3
Look
at
the
Exponent
91
3.1
Introduction
91
3.2 Local-global principle 92
3.3 Legendre's formula . . 96
3.4
Problems
with
combinatorial
and
valuation-theoretic aspects 104
3.5 Lifting
exponent

lemma
110
3.6 p-adic techniques . . . . 116
3.7 Miscellaneous problems 125
3.8 Notes

134
Addendum
3.A Classical
Estimates
on
Prime
Numbers 135
Addendum
3.B
An
Introduction
to
p-adic
Numbers
141
4
Primes
and
Squares
4.1 Notes

.
189
203

xvi
5 T
2
's
Lemma
5.1 Notes . .

.
Addendum
5.A Holder's Inequality in
Action.
6
Some
Classical
Problems
in
Extremal
Graph
Theory
6.1 Notes

.
Addendum
6.A Some Pearls of
Extremal
Graph
Theory.
Addendum
6.B Probabilities
in

Combinatorics

.
7
Complex
Combinatorics
7.1 Tiling
and
coloring problems
7.2
Counting
problems . . .
7.3 Miscellaneous problems

.
7.4 Notes

.
Addendum
7.A
Finite
Fourier Analysis
8
Formal
Series
Revisited
8.1
Counting
problems . .
8.2

Proving identities using generating functions
8.3 Recurrence relations

8.4 Additive properties . . .
8.5 Miscellaneous problems
8.6 Notes

.
Addendum
8.A Lagrange's Inversion
Theorem
9 A
Little
Introduction
to
Algebraic
Number
Theory
9.1 Tools from linear algebra
9.2 Cyclotomy

.
9.3
The
gcd trick

.
9.4
The
theorem

of symmetric
polynomials.
9.5 Ideal
theory
and
local
methods
9.6 Miscellaneous problems

.
9.7 Notes

.
Addendum
9.A
Equations
over
Finite
Fields
Addendum
9.B A Glimpse of Algebraic
Number
Theory
Contents
205
226
. 227
235
251
252

265
281
281
285
299
309
310
337
339
349
354
361
371
375
377
395
396
402
408
410
420
426
433
434
456
Contents
10
Arithmetic
Properties
of

Polynomials
10.1
The
a - blf(a) - f(b) trick

.
10.2 Derivatives
and
p-adic Taylor
expansions.
.
10.3 Hilbert polynomials
and
Mahler expansions
10.4 p-adic
estimates

10.5 Miscellaneous problems
10.6 Notes

.
11
Lagrange
Interpolation
Formula
11.1 Notes

.
12
Higher

Algebra
in
Combinatorics
12.1
The
determinant
trick

.
12.2 Matrices over
lF2

.
12.3 Applications of bilinear algebra
12.4
Matrix
equations

.
12.5
The
linear independence trick
12.6 Applications
to
geometry
12.7 Notes

.
Bibliography
xvii

485
485
494
497
506
513
520
521
537
539
541
546
552
561
568
576
583
585
Chapter
1
Some
Useful
Substitutions
Let
us first recall
the
classical
substitutions
that
will

be
used
in
the
fol-
lowing problems. All
of
these
are
discussed
in
detail
in
[3],
chapter
1
and
the
reader
is
invited
to
take
a closer look
there.
Consider
three
positive real
numbers
a,

b,
c.
If
abc
= 1, a classical
substi-
tution
is
x
a=
-
,
y
b=
~
z
,
z
c=

x
A less classical one is
a=_x_,
b=-Y-
c=_z_
y+z
z+x'
x+y
(for
some

positive
real
numbers
x,
y,
z)
whenever
ab
+
bc
+ ca + 2abc =
1,
or
its
equivalent version
y+z
a=
x '
b=Z+x
Y ,
x+y
c=
z
when
abc
= a + b + c + 2
(the
equivalence
between
the

two
relations
follows
from
the
substitution
(a,
b,
c)
t
(~,
i,
~)).
Two
other
very
useful
substitutions
concern
the
relations
a
2
+ b
2
+ c
2
=
abc
+ 4

and
a
2
+ b
2
+ c
2
+ 2abc =
1.
In
the
first case,
with
the
extra
assumption
max(a,
b,
c)
2:
2"
one
can
find positive
numbers
x, y, Z
such
that
xyz
= 1

and
111
a=x+-,
b=y+-,
c=z+
x y z
2
Chapter 1. Some Useful Substitutions
In
the
second
case
one
can
find
an
acute-angled
triangle
ABC
such
that
a =
cos(A),
b =
cos(B),
e =
cos(C).
Of
course,
in

practice
one
often
needs
to
use
a
mixture
of
these
substitutions
and
to
be
rather
familiar
with
classical
identities
and
inequalities.
But
expe-
rience
comes
with
practice,
so
let
us

delve
into
some
exercises
and
problems
to
see
how
things
really
work.
1.1
The
relation
a
2
+ b
2
+ c
2
=
abc
+ 4
We
start
with
an
easy
exercise,

based
on
the
resolution
of
a
quadratic
equation.
1.
Prove
that
if
a,
b,
e 2 0
satisfy
la
2
+ b
2
+ c
2
-
41
=
abc,
then
(a
- 2)(b -
2)

+
(b
- 2)(c -
2)
+ (c - 2)(a -
2)
2
o.
Titu
Andreescu,
Gazeta
Matematidi
Proof.
If
max(a,
b,
c) < 2,
then
everything
is
clear,
so
assume
that
max(a,
b,
c) 2
2.
Then
a

2
+ b
2
+ c
2
-
abc
= 4,
so
there
exist
positive
numbers
x, y, z
such
that
xyz
= 1
and
1
a =
x+-,
x
1
b = y
+-,
y
But
then
a,

b,
e
22
and
we
are
done
again.
1
c = z
+
z
o
Proof.
The
most
natural
idea
is
to
consider
the
hypothesis
as
a
quadratic
equation
in
a,
for

instance.
It
becomes
a
2
±
abc
+ b
2
+ c
2
-
4 = 0,
and
solving
the
equation
yields
=Fbe
±
J(b
2
- 4)(c
2
-
4)
a = 2 .
1.1. The relation a
2
+ b

2
+ e
2
=
abc
+ 4
Thus
(b
2
- 4)(e
2
-
4) =
(be
± 2a)2,
which
can
also
be
written
as
(b
- 2)(e -
2)
=
(be
±
2a)2
>
o.

(b+2)(c+2)
-
3
Writing
similar
expressions
for
the
other
two
variables,
we
are
done.
0
The
following exercise is
trickier
and
one
needs
some
algebraic
skills
in
order
to
solve
it.
We

present
two
solutions,
neither
of
which
is
really
easy.
2.
Find
all
triples
x,
y, z
of
positive
real
numbers
such
that
{
x2
+
y2
+
z2
=
xy
z + 4

xy
-+-
yz
+
zx
= 2(x + y + z)
Proof.
By
the
second
equation
we
have
max(x,
y,
z)
2 2
and
so
the
first
equa-
tion
yields
the
existence
of
positive
real
numbers

a,
b,
c
such
that
and
abc
= 1.
1
x =
a+-,
a
Since
abc
= 1, we
have
so
the
second
equation
can
be
written
The
left-hand
side
is
also
equal
to

1
z=c+
-
c
4
Chapter 1.
Some
Useful Substitutions
because
~
~
_ a
2
+ b
2
_
(2
b2)
b + a -
ab
- e a + .
We deduce
that
CL
a-I)
(2:::
ab
- 1) =
4.
Since

2:::
a
2':
3
and
2:::
ab
2':
3 (by
the
AM-GM
inequality
and
the
fact
that
abc
= 1),
this
can
only
happen
if
a = b = e = 1
and
thus
when x = y = Z =
2.
0
Proof.

If
x + y =
2,
the
second
equation
yields
xy
=
4,
so
that
(x
- y)2 =
-12
which is a contradiction.
Thus
x + y
=f
2
and
similarly y + z
=f
2, z + x
=f
2.
The
second
equation
yields

4-
xy
z=2+
x
+y
-
2'
and
a
rather
brutal
insertion
of
this
expression
in
the
first
equation
gives
(X_y)2+(
4-X
y
)2
x+y-2
(4
- xy)2
2-x-y
Unless x = y =
2,

this
implies
the
inequality 2 > x +
y.
If
two of
the
numbers
x,
y,
z are equal
to
2,
then
trivially so is
the
third
one.
If
not,
the
previous
argument
shows
that
2 > x +
y,
2>
y + z

and
2 > x + z.
But
then
the
second
equation
yields
x+y+z>
LX(Y;z)
=xy+yz+zx=2(x+y+z),
eye
a contradiction.
Thus,
the
only solution
is
x = Y = z =
2.
o
The
following problem hides
under
a clever algebraic
manipulation
a very
simple
AM-GM
argument.
The

inequality is
quite
strong, as
the
reader
can
easily see
by
trying
a brute-force approach.
3.
Prove
that
if a,
b,
e
2':
2 satisfy a
2
+ b
2
+ e
2
= abc + 4,
then
a + b + e +
ab
+ ae +
be
2':

2J(a
+ b + e + 3)(a
2
+ b
2
+ e
2
-
3).
Marian
Tetiva
1.1. The relation a
2
+ b
2
+ e
2
= abc + 4
5
Proof.
The
hypothesis yields
the
existence of positive real numbers
x,
y, Z such
that
x y
a = -
+-,

y x
b
= Y +
.:,
z y
x z
e = -
+
z x
The
miracle is
that
both
sides of
the
inequality have very nice factorizations.
For
the
right-hand
side,
this
is easy
to
observe, since
a+b+e+3=(xy+yz+zx)
-+-+-
(
1 1 1 )
xy
yz

zx
and
For
the
left-hand side, things are more subtle,
but
one finally reaches
the
identity
The
desired inequality becomes simply
the
AM-GM
inequality for two num-
bers! 0
The
following problems are
rather
tricky
mixtures
of
algebraic manipula-
tions
and
elementary
number
theory.
4.
Find
all

triplets
of
positive integers
(k,
l,
m)
with
sum
2002
and
for which
the
system
has real solutions.
x Y
-+-=k
Y x
z x
-+
=m
x z
Titu
Andreescu, proposed for IMO 2002
6
Chapter
1.
Some
Useful
Substitutions
Proof.

The
system
has
solutions if
and
only if
k
2
+
l2
+ m
2
=
lkm
+
4.
An
easy
computation
shows
that
this
relation is equivalent
to
(k
+
2)(l
+
2)(m
+ 2) =

(k
+ l + m + 2)2.
As k + l + m = 2002,
we
deduce
that
any
solution
of
the
problem satisfies
k + l + m = 2002
and
(k
+
2)(l
+
2)(m
+
2)
=
(k
+ l + m + 2)2 = 2004
2
=
24
. 3
2
.167
2

.
A simple case analysis shows
that
the
only solutions are k = l = 1000, m = 2
and
its
permutations.
The
result follows. 0
5.
Solve
in
positive integers
the
equation
(x+2)(y+2)(z+2)
=
(x+y+z+2)2.
Titu
Andreescu
Proof. A simple algebraic
manipulation
shows
that
the
equation
is equivalent
to
x2+y2+z2

=
xyz+4
and,
seeing
this
as a
quadratic
equation
in
z,
we
obtain
the
equivalent form
(x2-4)(y2-4)
=
(xy_2z)2.
If
x2 < 4,
then
y2
< 4 as well
and
so x = y =
1,
yielding z =
2.
If
x =
2,

then
y = z.
In
all
other
cases,
we
can
find a positive square-free integer D (which is easily seen
to
be
different
from
1)
and
positive integers u, v such
that
x2 - 4 =
Du
2
and
y2
- 4 =
Dv
2
.
Thus, solving
the
problem comes down
to

solving
the
generalized Pell
equation
a
2
-
Db
2
=
4,
which is a classical topic:
this
equation
always has nontrivial
integer solutions
and
if (ao,
bo)
is
the
smallest solution
with
ao,
bo
> 0,
then
all solutions are given by
1.1.
The

relation a
2
+ b
2
+ c
2
= abc + 4
7
~.
~
Jv
[(no
+
;vn)
n _ (
aD
-
;vn)
nJ.
o
Part
of
the
following
problem
can
be
dealt
in
a classical way,

but
we do
not
know how
to
solve
it
entirely
without
using
the
trick of
substitutions.
6.
The
sequence (a
n
)n2:0
is defined
by
ao
=
al
= 97
and
an+l =
anan-l
+
J(a~
-

l)(a~_l
-
1)
for all n
2:
1.
Prove
that
2 +
y2
+ 2a
n
is a perfect
square
for all
n.
Proof. Writing
(an+! - a
n
a
n
_d
2
=
(a;
-
l)(a;_l
-
1)
and

simplifying
this
expression yields
2 2 2 2 1
an-l
+ an + an+l -
anan-lan+l
= ,
thus
Titu
Andreescu
(2a
n
_l)2
+ (2a
n
)2
+ (2a
n
+d
2
-
(2a
n
-l)(2a
n
)(2a
n
+d
= 4.

Since
we
clearly have an > 2 for all n,
this
implies
the
existence
of
a sequence
x
n
> 1 such
that
2a
n
=
Xn
+
x;:;-l
and
such
that
Xn+l =
XnXn-l.
Thus
logxn
satisfies a Fibonacci-type recursive relation
and
so we
can

immediately find
out
the
general
term
of
the
sequence (an)n. Namely, a small
computation
shows
that
if
we
define a = 2 +
yI3,
then
Xn
= a
4Fn
, where Fn is
the
nth
Fibonacci number.
Thus
1(
4F
1)
an
="2
a n + a

4Fn
and
so
2 +
y2
+
2a
n
= 2 + ( a
2Fn
+ a
21Fn
) =
(a
Fn
+
a~n)
2
The
result follows, since
an
+
a-
n
E Z for all
n,
by
the
binomial formula. 0
S

Chapter 1.
Some
Useful
Substitutions
Remark
1.1.
Here
is
an
alternative
proof
of
the
fact
that
all
terms
of
the
sequence
are
integers,
without
the
use
of
substitutions.
The
method
that

we
will use for
this
problem
appears
in
many
other
problems.
As
we
saw
in
the
previous
solution,
the
sequence satisfies
the
recursive
relation
Writing
the
same
relation
for n + 1
instead
of
n
and

subtracting
the
two yields
the
identity
a~+2
-
a~-I
= 2a
n
a
n
+l(a
n
+2
-
an-I).
Note
that
(an)n is
an
increasing sequence
(this
follows
trivially
by
induction
from
the
recursive

relation),
so
that
we
can
divide
by
a
n
+2
-
an-I
i-
0
in
the
previous
relation
and
get a
n
+2
= 2a
n
a
n
+l -
an-I.
The
last

relation
clearly
implies
that
all
terms
of
the
sequence
are
integers (since
one
can
immediately
check
that
this
is
the
case
with
the
first
three
terms
of
the
sequence).
Note
however

that
it
does
not
seem
to
follow easily
that
2 +
V2
+ 2a
n
is a
perfect
square
using
this
method.
Remark
1.2.
There
are
a
lot
of
examples
of
very
complicated
recurrence

relations
that
rather
unexpectedly
yield integers.
For
instance,
the
reader
can
try
to
prove
the
following
result
concerning
Somos-5 sequences: let
al
=
a2
=

=
a5
= 1
and
let
for n
2:

O.
Then
an is
an
integer
for all
n.
Similarly
one
defines Somos-6,
Somos-7,
etc
sequences
by
the
formulas
ao
=
al
=

=
a5
= 1,
ao
=
al
=

=

a6
= 1,
etc.
One
can
prove
(though
this
is
not
easy)
that
all
terms
of
Somos-6
and
Somos-7 sequences
are
integers. Surprisingly,
this
fails for Somos-S sequences
(in
which
case
al7
is
no
longer
an

integer!).
1.2.
The
relations abc = a + b + e + 2 and'ab +
be
+ ea + 2abe = 1
1.2
The
relations
abc
= a + b + e + 2
and
ab
+
be
+
ea
+ 2abe = 1
9
The
first
inequality
in
the
following
problem
is very useful
in
practice
and

we will
meet
it
very
often
in
the
following
problems.
7.
Prove
that
if
x,
y, z > 0
and
xyz
= x + y + z + 2,
then
3
xy
+
yz
+
zx
2:
2(x
+ y +
z)
and

Vx
+
vY
+
vZ
::;
'2vxyz.
Proof.
With
the
usual
substitutions
b+e
e+a
x=
y-
a'
-
-b-'
a+b
z=
e '
the
first
inequality
comes
down
(after
clearing
denominators

and
canceling
out
similar
terms)
to
Schur's
inequality
a(a
- b)(a -
c)
+ b(b - a)(b -
c)
+ e(e -
a)(e
-
b)
2:
0,
while
the
second
one
follows.
by
adding
up
the
inequalities
o

The
reader
may
find a
bit
strange
the
first
method
of
proof
of
the
following
problem,
but
it
is
actually
a
quite
powerful one. We will use
again
this
kind
of
argument,
see
problem
11

for
instance.
Also,
the
third
solution
uses a very
useful
technique.
S.
Let
x,
y, z > 0
be
such
that
xy
+
yz
+
zx
=
2(x
+ y +
z).
Prove
that
xyz
::;
x + y + z + 2.

Gabriel
Dospinescu,
Mircea
Lascu
10
Chapter 1. Some Useful Substitutions
Proof. We will argue
by
contradiction, assuming
that
xyz
> x + y + z +
2.
We
claim
that
we
can
find 0 < r < 1 such
that
X =
rx,
Y = ry, Z = r z satisfy
XY
Z = X + Y + Z +
2.
Indeed,
this
comes down
to

the
vanishing of
f (r) =
r3
xy
z - r (x + y + z) - 2
between 0
and
1,
and
this is clear, since f(O) < 0
and
f(1)
>
O.
Next,
the
condition
xy
+
yz
+
zx
=
2(x
+ y + z) yields
XY
+
YZ
+

ZX
=
2r(X
+ Y +
Z)
<
2(X
+ Y +
Z).
This
contradicts
the
first inequality of problem
7.
o
Proof.
The
condition
can
also
be
rewritten
in
the
form
(x - 1)(y - 1) + (y - 1)(z - 1) + (z -
1)(x
- 1) = 3
or
in

the
form
xyz
- x - y - z - 1 = (x - 1)(y - 1)(z - 1).
We will discuss several cases.
If
x, y, z
:::::
1,
then
by
the
A~\1-GM
inequality
and
the
first identity,
we
get
which yields,
thanks
to
the
second identity,
the
desired estimate.
If
x,
y,
z

::;
1 or if only one
of
the
numbers
x,
y, z is smaller
than
or
equal
to
1,
then
(x - 1)(y - 1)(z -
1)
::;
0
and
so
xyz
::;
x + y + z + 1
in
this
case.
Finally, if two of
the
numbers are smaller
than
1,

say
x,
y
::;
1,
the
desired
inequality
can
be
written
in
the
form 0
::;
x + y + z(1 -
xy)
+ 2, which is
obvious.
0
Proof. For
three
positive real numbers x, y, z consider fixing
the
first two el-
ementary
symmetric polynomials
171
= x + y + z
and

172
=
xy
+
yz
+
zx
and
letting
173 =
xyz
vary.
This
amounts
to
varying only
the
constant
term
in
the
polynomial
1.2. The relations
abc
= a + b + e + 2 and,ab +
be
+ ea + 2abe = 1
11
and
defining

X,y,z
to
be
the
three
roots
of
this
polynomial (in some order).
Increasing
173, i.e. lowering
the
constant
term,
corresponds geometrically
to
lowering
the
graph. As we lower
the
graph,
the
smallest
root
increases,
thus
we
maintain
three
positive real

roots
until
the
smallest
root
becomes a double
root.
If
the
double
root
is
at
t = a
and
the
larger
root
at
t =
b,
then
we
have
171
= 2a +
b,
172 = a
2
+ 2ab

and
17
3 = a
2
b.
If
we
fix
171
and
172
with
172 = 2171 as hypothesized
then
we
find b =
~
,
2(a-1
J
and
because 0 < a
::;
b,
we
see
that
1 < a
::;
2.

By
the
discussion above
xyz
= 173
::;
a
2
b,
so
it
suffices
to
show
that
for 1 < a::; 2.
But
this
rearranges
to
(a-2)2(a
2
-1)
:::::
0
and
we
are
done. 0
The

technique used
in
the
first solution of
the
next
problem is
rather
ver-
satile
and
the
reader
is invited
to
read
the
addendum
5.A for more examples.
9.
Let
x,
y, z > 0
be
such
that
xy
+
yz
+

zx
+
xyz
=
4.
Prove
that
Proof. Using
the
usual
substitution
2a
2b
x =
b+e'
y=e+a'
the
problem reduces
to
proving
the
inequality
2e
Z=
a+b'
Gabriel Dospinescu
3
b+e
> 16
(a+b+e)3

(
)
2
L Iff! -
(a
+
b)(b
+ e)(e + a)'
12
Chapter 1.
Some
Useful Substitutions
This
is a
quite
strong
inequality
and
it
is easy
to
convince oneself
that
most
applications
of
classical techniques fail. However,
the
following
smart

application of Holder's inequality does
the
job:
so
it
is enough
to
prove
that
This
reduces
after
expanding
to
2:
a(b -
e)2
2':
0, which is clear. 0
Proof.
First,
we get
rid
of
those
nasty
square
roots,
via
the

substitution
4
2 4 2
zx
= 4b
2
•
xy
=
e,
yz
=
a,
Then
2be
2ea 2ab
x=~,
y=b'
z=~
and
replacing these values in
the
inequality yields
the
equivalent form
3(a
+ b +
e)2
2':
16(a + be)(b + ea)(e + ab).

The
hypothesis becomes a
2
+ b
2
+ e
2
+ 2abe = 1, so
that
there
exists
an
acute-
angled
triangle
ABC
such
that
a =
cosA,
b =
cosB,
e =
cosC.
Next,
observe
that
e +
ab
=

cosC
+ cos A .
cosB
= - cos(A +
B)
+ cos A .
cosB
= sin A .
sinB.
Using
this
(and
similar identities
obtained
by
permuting
the
variables),
the
desired inequality becomes
v'3
L cos A
2':
4
II
sin A.
Using
the
well-known identities
s

= 4R
II
cos
~
, r = 4R
II
sin
~
, L cos A = 1 +
~,
1.2. The relations abc = a + b + e + 2 and'ab +
be
+ ea + 2abe = 1
the
inequality becomes
v'3r+R>
2rs
R -
R2'
13
or
(R+r)Rv'3
2':
2rs.
This
splits trivially
into
R+r
2':
3r

(Euler's inequality)
and
s
::;
3r
R,
the
last
one being well-known
and
easy. 0
Here is yet
another
easy
application
of
problem
7.
10.
Let
u, v, w > 0
be
positive real
numbers
such
that
u + v + w +
';uvw
=
4.

Prove
that
fUV
fvW
fWU
V
;;;
+ V
;;
+ V
;;
2':
u + v + w.
China
TST
2007
Proof. To get
rid
of
those
nasty
square
roots,
let
us
perform
the
substitution
~=e,
~=a,

(f-=b.
Then
u =
be,
v = ea
and
w =
ab,
so
that
the
inequality becomes a + b + e
2':
ab
+
be
+ ea
and
the
hypothesis is
ab
+
be
+ ea + abc =
4.
Another
substitution
a
=~,
b -

~
e =
~
x -
y'
z
yields
xyz
=
x+y+z+2.
We need
to
prove
that
xy+yz+zx
2':
2(x+y+z),
which is
the
first
component
of problem
7.
0
The
following problem
has
some
common
points

with
problem 7
and
one
can
actually deduce it from
that
result.
But
the
proof
is
not
formal.
11. Prove
that
if
a,
b,
e > 0
and
x = a +
-b
1
, y
= b +
~,
z = e +
~,
then

e a
xy+yz+zx
2':
2(x+y+z).
Vasile
Cartoaje
14
Chapter 1.
Some
Useful
Substitutions
Proof.
The
method
is
the
same
as
the
one used
in
problem
8.
The
starting
point
is
the
observation
that

xyz
2:
2 + x + y + z. Indeed,
1 1 1 1
.
xyz
= abc + - + a + b + e + - +
-b
+ -
2:
2 + x + y +
z.
abc a e
Let
us assume for sake of contradiction
that
xy
+
yz
+
zx
<
2(x
+ y +
z)
and
let us choose r E (0,1] such
that
X =
rx,

Y =
ry,
Z =
rz
satisfy
XY
Z =
2 + X + Y +
Z.
This
equality is equivalent
to
r3
xy
z = 2 + r
(x
+ y +
z)
and
such
r exists by continuity
of
the
function
f(r)
= r
3
xyz-2-r(x+y+z)
and
by

the
fact
that
f(I)
2:
0
and
f(O) <
O.
The
hypothesis
xy
+
yz
+
zx
<
2(x
+ y +
z)
can
also
be
written
as
XY
+
YZ
+
ZX

<
2r(X
+ Y +
Z)
~
2(X
+ Y +
Z).
This
contradicts
the
first inequality
in
problem
7.
o
Proof. As
in
the
previous proof,
we
obtain
that
xyz
2:
x + y + z +
2.
Since we
obviously have
min(xy,

yz,
zx)
> 1,
this
implies
that
z
2:
2;:yX_+l
and
similar
inequalities
obtained
by
permuting
the
variables. Next,
we
have
1 1 1
x + y + z = a + - + b + - + e + - >
6.
abc
-
In
particular,
there
are
two numbers, say
x,

y, such
that
x + y
2:
4.
The
inequality
to
be
proved is equivalent
to
z
2:
2(~~~~2XY,
so we are done if
we
can
prove
that
2 + x + Y
2(x
+ y) -
xy
~
> .
xy-I
-
x+y-2
But
this

is equivalent,
after
an
easy
computation
(in which
it
is convenient
to
denote
S = x + y
and
P =
XV),
to
(xy
- y - x)2
2:
(x
-
2)(y
- 2).
If
(x
-
2)(y
-
2)
~
0,

this
is clear; otherwise
x,
y
2:
2 as x + y
2:
4.
Let
u = x -
2,v
= y -
2,
then
the
inequality becomes (uv + u +
v?
2:
uv,
with
u, v
2:
0
and
it
is obvious. 0
The
form
of
the

following inequality strongly suggests
the
use of
the
Cauchy-Schwarz inequality.
It
turns
out
that
this
approach
works,
but
in
a
rather
indirect
and
mysterious way, which makes
the
problem
rather
hard.
1.2. The relations abc = a + b + e + 2
and
tlb +
be
+ ea + 2abe = 1
15
12. Prove

that
for all a,
b,
e > 0,
Japan
1997
Proof.
With
the
usual
substitution
b+e
a+e
x =
-a-'
y =
-b-'
a+b
z= ,
e
the
problem asks
to
prove
that
for
any
positive
numbers
x,

y,
z such
that
xy
z = x + y + z + 2
we
have
(x-I)2
(y
I?
(z-I)2
3
-' , "
+ + >

x
2
+ 1
y2
+ 1
z2
+ 1 - 5
Applying
the
Cauchy-Schwarz inequality (or
what
is also called
Titu's
lemma),
we

obtain
the
bound
(x
1)2
(Y-I?
(z
1)2
(x+y+z
3)2
x
2
+ 1 +
y2
+ 1 +
z2
+ 1
2:
x
2
+
y2
+ z2 + 3 .
It
remains
to
prove
that
the
last

quantity
is
at
least
~.
Let
S = x + y + z
and
P =
xy
+
yz
+
zx,
so
that
the
inequality becomes
Notice
that
as % +
~
2:
2
and
cyclic
permutations
of this,
we
have S

2:
6.
Also,
by problem 7
we
have P
2:
2S.
Therefore,
(S
- 3)2 >
(S
3)2 = S - 3 = 1
__
2_
>
~.
S2 -
2P
+ 3 - S2 -
4S
+ 3 S - 1 S - 1 - 5
o
Remark
1.3.
There
are
other
solutions for
this

problem: some
of
them
are
shorter,
but
not
easy
to
find. Here is a
particularly
elegant one, based
on
16
Chapter
1.
Some
Useful
Substitutions
the
linearization
technique:
the
inequality
is
homogeneous,
so we
may
assume
that

a + b + e = 1.
We
need
to
prove
that
The
point
is
to
bound
from
below
(
(1-~t)2
2
by
an
affine
function
of
a,
suitably
I-a
+a
chosen.
The
best
choice is
the

following
Of
course,
the
question
is
how
one
came
up
with
something
like
this.
Well,
it
is
actually
very
easy: we
need
constants
A,
B
such
that
for all a E
[0,
1],
with

equality
for a =
~.
Impos~ng
also
th~
vanisfi~g
of
the
derivative
of
the
difference
between
the
left
and
nght-
hand
SIde
at
3"
YIelds
the
desired
constants
A, B.
Once
we
have

the
previous
estimates,
it
is
very
easy
to
conclude
by
adding
them
and
taking
into
account
that
a + b + e =
1.
13.
Find
all
real
numbers
k
with
the
following
property:
for all

positive
numbers
a,
b,
e
the
following
inequality
holds
Vietnam
TST
2009
Proof.
Take
first
of
all a = b = 1
and
arbitrary
e
to
get
that
1.2.
The
relations abc = a + b + e + 2
and
ab +
be
+ ea + 2abe = 1

17
Letting
e -+ 0, we
deduce
that
any
k
as
in
the
statement
must
satisfy
k(k+1)22:
(k+~)3
which
is
equivalent
to
4k2 + 2k
2:
1.
We
claim
that
any
such
k is a
solution
of

the
problem.
Pick
such
k
and
perform
the
usual
substitution
a
X=
b+e'
to
reduce
the
problem
to
b
y=e+a'
e
Z=
a+b
k
3
+ k
2
(x
+ y + z) +
k(xy

+
yz
+
zx)
+
xyz
2:
(k
+~)
3
Now, we
know
that
xy
+
yz
+
zx
+
2xyz
= 1
and
it
is
by
now
a classical
fact
(use
problem

7 for
l/x,l/y,l/z)
that
x+y+z
2:
2(xy+yz+zx).
Thus,
it
is
enough
to
ensure
that
k
3
+ (2k2 +
k)(xy
+
yz
+
zx)
+ 1 -
(xy
~
yz
+
zx)
2:
(k
+~)

3
This
can
be
rewritten
as
(2k2
+ k -
~)
(Xy
+
yz
+
zx
-
i)
2:
o.
But
the
last
inequality
follows from
the
fact
that
2k2 + k -
1.
> 0
and

2 -
3
xy
+
yz
+
zx
2:
4'
which,
by
returning
to
the
substitution
b
e
a
X=
b+e'
y= ,
e+a
z = a +
b'
is
equivalent
to
L, a(b -
e)2
2:

o.
In
condusion,
the
solutions
of
the
problem
are
the
real
numbers
k
such
that
4k2 + 2k
2:
1. 0
18
Chapter 1. Some Useful Substitutions
We
end
this
section
with
a very challenging problem, which answers
the
following
natural
question:

what
would
be
the
analogue
of
the
classical sub-
stitution
x =
b!e,
y =
eta,
Z =
a~b
when
we have five variables? We
warn
the
reader
that
the
first solution is really
not
natural.

14. Let
aI,
a2,

,
a5
be
positive real
numbers
such
that
What
is
the
least possible value of
a\
+
:2
+ +
a5
Gabriel Dospinescu,
Mathematical
Reflections
Proof. Consider
the
linear
system
We
can
try
to
solve
this
system

by
expressing, for example, all variables
in
terms
of
Xl,
X5.
Namely, we
can
use
the
fifth, first
and
second
equation
to
express
X2, X3,
X4
in
terms
of
Xl,
X5.
Replacing
the
obtained
values
in
the

other
two
equations
and
eliminating
Xl,
x5
between
them,
we
obtain
that
the
system
has
a nontrivial solution
if
and
only if
5 5
II
ai
= 2 + L
ai
+
ala4
+
a4
a
2

+
a2
a
5
+
a5
a
3
+
a3
a
l·
i=l
i=l
All
the
previous (painful!)
computations
are
left
to
the
reader, since
they
are
far from having
anything
conceptual.
Note
that

the
same
result
can
be
obtained
by
computing
the
determinant
of
the
associated
matrix.
Now,
the
previous relation is almost
the
one given in
the
statement,
up
to
a
permutation
of
the
variables. So, since
the
conclusion is

symmetric
in
the
five variables, we
will assume from now
on
that
the
previous
relation
is satisfied
instead
of
the
one given
in
the
statement.
1.2. The relations abc = a + b + e + 2 and
ab
+
be
+ ea + 2abe = 1
19
The
crucial claim is
that
under
the
previous hypothesis,

the
system
has
solutions whose unknowns
Xi
are
positive.
Probably
the
easiest way
to
prove
this
is
to
exhibit
such a solution. Well, simply
take
Xl
=
1,
1 +
al
+
a2
+
a3
+
ala3
X5

=
= = =-:::.
1 +
a5
+
a3a5
+
a2
a
5
and
the
define
1 +
X5
X4
+
X5 X3
+
X4
X4
=
,
X3
=
X2
=
= ~
a4
a3

a2
The
question is,
of
course, how
on
earth
did we choose
the
value
of
X5?
Well,
simply by solving
the
system, as
indicated
in
the
beginning
of
the
solution.
Note
that
we clearly have
Xi
> 0
and
an

easy
computation
shows
that
these
are
solutions
of
the
system
(we only have
to
check two equations, since
three
are
satisfied
by
construction).
The
conclusion is
that
if
the
ai's
satisfy
the
condition
5 5
II
ai

= 2 + L
ai
+
ala4
+
a4
a
2
+
a2a5
+
a5a3
+
a3
a
l,
i=l i=l
then
we
can
find
Xi
> 0 such
that
X3
+
X4
Xl
+
X2

a2
= ,

,a5
=
-" =-
X2 X5
So,
the
problem
reduces
to
finding
the
minimal
value
of
Xl
X2
X5
+
+
+
X2
+
X3 X3
+
X4
Xl
+

X2
We claim
that
the
previous expression is always
at
least
5/2.
By
Titu's
lemma,
this
reduces
to
proving
that
But
since
(Xl
+
X2
+
X3
+
X4
+
X5)2
2:
~
L

XiXj.
i<j
20
Chapter 1. Some Useful Substitutions
this
reduces
to
(~Xi)2
::; 5 ~
x;,
which is Cauchy-Schwarz.
Putting
everything together,
we
obtain
that
the
minimal value of
is
5/2,
attained
when all
ai's
are equal
to
2.
Who
wants
to
play now

the
same
game
with
7 variables? 0
Proof. Here is
another
proof, also far from being evident. " We will use
the
following nontrivial
Lemma
1.4.
For all nonnegative real numbers
Xl,
X2,

,X5
we have
(Xl
+ X2 + +
X5)3
:2:
25(XIX2X3
+
X2X3X4
+ +
X5XIX2).
Proof.
This
is

not
easy
at
all,
but
here is a very elegant
(but
unnatural)
proof:
consider
the
identity
XIX2X3
+
X2X3X4
+
X3X4X5
+
X4X5XI
+
X5
X
I
X
2
=
X5(XI
+
X3)(X2
+

X4)
+
X2X3(XI
+
X4
X5).
We
may
assume
that
X5
=
min
Xi
(since
the
inequality is cyclic), so
that
Xl
+
X4
-
X5
:2:
O.
Denoting
Xl
+
X2
+

X3
+
X4
=
4t
and
using
the
AM-GM
inequality, we deduce
that
Thus,
it remains
to
prove
that
2
(4t-
X
5)3
(X5+
4t
)3
4t
X5
+ 3
::;
25
By
homogeneity,

we
may
assume
that
X5
= 1
and
then
expanding
everything
the
inequality becomes,
with
the
substitution
4t
1 = 3x,
(x-1)2(8x+7):2:
0,
which is clear. 0
1.3. The relation a
2
+ b
2
+ c
2
+
2abc
= 1
Using

this
lemma
for
the
inverses of
the
ai's
and
denoting
we
deduce
that
111
S=-+-+

+_,
al
a2
a5
21
On
the
other
hand,
by Mac-Laurin's
and
AM-GM
inequalities we also have
Taking into account these inequalities
and

the
hypothesis,
we
deduce
that
which immediately implies
that
S
:2:
~.
o
1.3
The
relation
a
2
+ b
2
+ c
2
+ 2abc = 1
The
following problem is a
rather
tricky application of
the
AM-GM in-
equality.
15.
Prove

that
in all acute-angled triangles
the
following inequality holds
(
COS
A)
2
(cos
B)
2
(cos
C)
2
cosB
+
cosC
+
cosA
+
8cosAcosBcosC:2:
4.
Titu
Andreescu,
MOSP
2000
Proof. Since
cos
2
A + cos

2
B + cos
2
C + 2 cos A cos B cos C = 1
,
22
Chapter 1.
Some
Useful
Substitutions
the
inequality
can
be
written
(
COSA)2
(COSB)2
(cosC)2
.

+
-C
+
-A
2:
4(cos
2
A+cos
2

B+cos
2
C).
cosB
cos cos
Let
x = cos
2
A, y = cos
2
B,
z = cos
2
C,
so
we
need
to
prove
that
x y z
-+-+
2:
4
(x+y+z).
Y z x
The
point
is
that

we actually have
x y z
x+y+z
-+-+->
y z x -
ijxyz
for
any
positive real numbers x,
y,
z, as follows by
adding
up
the
AM-GM
inequalities
2x
+
'!{
2:
3
3
(;2.
y z V
Yz
Thus,
it
is enough
to
prove

that
xyz
:S
t4'
But
this
is
well-known
and
easy
to
prove. 0
The
following problem is a
preparation
for a
hard
problem
to
come,
but
has also
an
independent
interest.
16. Prove
that
in
every acute-angled triangle
ABC,

(cos A + cosB)2 +
(cosB
+
cosC)2
+
(cosC
+ cosA)2
:S
3.
Proof.
Letting
cos A =
0,
YYz
cosB
=
0,
Y-;;
cosC
= 0
Y-;;Y
1.3. The relation a
2
+ b
2
+ c
2
+ 2abc = 1
23
for some positive real x,

y,
z, we
obtain
the
relation
xyz
= x + y + z + 2 from
the
classical
identity
L cos
2
A + 2 n cos A =
1.
Thus
we
can
find positive real
numbers
a,
b,
c such
that
b+c
X= ,
a
c+a
y=
-b-'
The

desired inequality
can
be
written
a+b
Z=
c
1 1 3 3
L + L
:S
-
-¢==}
LX+
LVxY:S
-xyz
xy
zyXY
2 2
-¢==}
LX
+ L
VxY
:S
~(x
+ y + z + 2)
-¢==}
(L
vxf
:S
2

(L
x +
3)
.
This
follows from Cauchy-Schwarz
o
Proof. Since
the
triangle is acute,
there
are
positive numbers
x,
y, z such
that
a
2
= y + z, b
2
= Z +
x,
c
2
= X +
y.
Then
x
cos A =
,.;===::;:=;=====c=

J(x
+ y)(x + z)
and
similar identities for cos
B,
cos
C.
The
desired inequality
can
be
written
as
x
2
yz
3
L (x + y)(x + z) + L
(y
+
z)J(x
+ y)(x + z)
:S
2'
which (after clearing denominators) is equivalent
to
L
(x
2
(y

+ z) +
yzJ(x
+ z)(x +
y))
:S
~(x
+ y)(y + z)(z + x)
and
then
to
1
L
Jyz(x
+
y)
.
yz(x
+ z)
:S
3xyz + 2
LYz(y
+ z).
eye
eye
24
Chapter
1.
Some
Useful
Substitutions

However,
this
follows
immediately
from
the
AM-GM
inequality:
1
Jyz(x
+
y)
.
yz(x
+
z)
:S
xyz
+ "2Yz(y +
z)
and
two similar inequalities.
D
Even
though
the
following
problem
seems classical,
it

is
actually
rather
hard.
Fortunately,
we
did
the
difficult
job
in
the
previous problem. We
also
present
an
independent
and
very elegant
approach
due
to
Oaz
Nair
and
Richard
Stong.
17.
Prove
that

if a,
b,
c 2 0 satisfy a
2
+ b
2
+ c
2
+
abc = 4,
then
o
:S
ab +
bc
+ ca - abc
:S
2.
Titu
Andreescu,
USAMO
2001
Proof.
The
inequality
on
the
left is easy:
the
hypothesis

and
the
AM-GM
inequality
imply
that
abc
:S
1,
so
that
min(a,
b,
c)
:S
1.
We
may
assume
that
c =
min(a,
b,
c),
so
that
ab+bc+ca
abc=c(a+b)+ab(l-c)20.
The
hard

point
is proving
that
ab +
bc
+
ca
- abc
:S
2.
Taking
into
account
the
hypothesis,
this
can
also
be
written
as
( a ;
b)
2 +
(b;
c)
2 + ( a ;
c)
2
:S

3.
If
we
denote
a =
2x,b
=
2y,c
=
2z
then
x
2
+
y2
+ z2 +
2xyz
= 1
and
so
there
exists a
triangle
ABC
such
that
x =
cos(A),
y =
cos(B),

z = cos(C).
Thus,
the
problem
reduces
to
the
previous one (since a,
b,
c
are
nonnegative,
the
triangle
ABC
is
acute
angled). D
Proof. Here is a
very
elegant
proof
for
the
hard
part
of
the
inequality.
Among

the
three
numbers
a-I,
b -
1,
c -
1,
two have
the
same
sign, say b - 1
and
c -
1.
Thus
(1
b)(1 -
c)
2 0, implying
that
b + c
:S
bc
+ 1
and
ab + ac - abc
:S
a.
1.3.

The
relation a
2
+ b
2
+ c
2
+
2abc = 1
25
Thus,
it
is
enough
to
prove
that
a +
bc
:S
2.
But
the
given
condition
and
the
fact
that
a is

nonnegative
imply
that
-bc
+
J(b
2
-
4)(c
2
-
4)
a=
~ ~ ~
2 .
Thus,
it
is
enough
to
prove
that
J(b
2
-
4)(c
2
- 4)
:S
4 -

bc.
Note
that
bc
:S
4,
since
b
2
:S
4
and
c
2
:S
4.
Squaring
the
previous
inequality
thus
yields
an
equivalent one
which is obvious.
D
We
end
this
chapter

with
another
challenging problem, which reduces
after
some
tricky
algebraic
manipulations
to
the
infamous
"Iran
1996 inequality."
18.
Prove
that
in all
acute-angled
triangles
the
following
inequality
holds
a
2
b
2
a
2
c

2
b
2
c
2
-2-+
+-2
29R2.
c a
Nguyen Son
Ha
Proof.
If
A,
B,
C
are
the
angles
of
the
triangle,
the
sine law
and
the
identity
sin
2
x + cos

2
X = 1 yield
the
equivalent form
of
the
inequality
Write
cos
2
A =
yz,
cos
2
B =
zx,
cos
2
C =
xy.
Then
xy
+
yz
+
zx
+
2xyz
= 1
and

so
there
exist
positive
numbers
X,
Y,
Z
such
that
X
Z
x = Y +
Z'
Y
y=
X+Z'
z=
X+Y'
We
want
to
prove
that
'"
' (
1
__
y
z-,-)-c-(

1_-_z_x-,-)
>
~.
L 1
xy
- 4
26
Chapter 1. Some Useful Substitutions
On
the
other
hand,
we have
X(X
+ Y +
Z)
1 -
yz
=
(X
+
Y)(X
+
Z)'
so
that
the
inequality
is equivalent
to

~
XY
(X
+ Y +
Z)
>
~
~
Z(X
+
y)2
-
4'
This
can
also
be
written
in
the
form
1 9
XYZ(X
+ Y +
Z)·
L
(ZX
+
ZY)2
2':

4
and
it
is a consequence
of
the
following famous
inequality
Lemma
1.5.
For all positive numbers a,
b,
e
we
have
(
1 1
1)
9
(ab
+
be
+ ea)
(b
+
e)2
+
(e
+
a)2

+
(a
+
b)2
2':
4'
Proof.
We
may
assume
that
a
2':
b
2':
e.
First,
we show
that
1 1 1 1 2
=-+ +
>-+
.
(a+b)2
(a+e)2
(b+e)2
-
4ab
(a+e)(b+e)
This

can
be
rewritten
(
1
1)2
(a
b)2
a + e - b + e
2':
4ab( a +
b)2
,
or
equivalently
4ab(a+b)2
2':
(a+e)2(b+e)2.
This
is clear, as (a+b)2
2':
(a+e)2
and
4ab
2':
(b
+
e)2.
Thus,
it

remains
to
prove
that
[
1
2]
9
(ab
+
be
+ ea)
4ab
+
(a
+ e)(b +
e)
2':
4'
1.4.
Notes
Using
the
identities
ab
+
be
+
ea
1 e( a +

b)
4ab
= 4 +
4ab
'
2(ab
+
be
+
ea)
2e
2
,'-_ , , _c-'-
= 2 -

__
_
(a+e)(b+e)
(a+e)(b+e)'
this
becomes
e(a +
b)
2e
2
~ '->
.
4ab
-
(a

+ e)(b +
e)
27
But
this
is
simply
the
standard
inequality
(a
+
b)
(a
+ e)
(b
+
e)
2':
8abe
and
we
are
done. 0
This
completes
the
solution
to
the

problem.
o
1.4
Notes
We would like
to
thank
the
following
people
for
their
solutions:
Vo
Quoc
Ba
Can
(problems
13, 18),
Xiangyi
Huang
(problems
4, 5),
Logeswaran
La-
janugen
(problem
1, 9),
Oaz
Nair

(problem
17),
Dusan
Sobot
(problems
2,
9,
16),
Richard
Stong
(problems
8, 17),
Gjergji
Zaimi
(problems
2,
3, 6,
7,
8, 11,
12, 15, 16, 17).
Chapter
2
Always
Cauchy-Schwarz

As
the
title
suggests, all problems in
this

chapter
can
be solved using
the
Cauchy-Schwarz inequality, even
though
sometimes
this
will require
quite
a
lot of work. Let us recall
the
statement
of
the
Cauchy-Schwarz inequality: if
aI,
a2,···,
an
and
b
l
,
b2,

, b
n
are
real numbers,

then
This
follows easily from
the
fact
that
L~=l
(aix
+ b
i
)2
:2:
0 for all real numbers
x. Indeed,
the
left-hand side is a
quadratic
function of x
with
nonnegative
values, so its discriminant
is
negative or
O.
But
this
is precisely
the
content of
the

Cauchy-Schwarz inequality.
Another
proof
is based
on
Lagrange's
identity
L (aibj - a
j
b
i
)2.
I
:Si<j:Sn
This
useful
identity
will be used several times
in
this
chapter.
As
the
best
way
to
get familiar
with
this
inequality is

via
a lot of examples
of all levels of difficulty,
we
will
not
insist on
any
theoretical aspects
and
go
directly
to
battle.
We
start
with
two easy examples, destined
to
give
the
reader
some confidence. He will surely need
it
for
the
more difficult problems
to
come

30
Chapter 2.
Always
Cauchy-Schwarz .

1.
Let
a,
b,
c
be
nonnegative
real
numbers.
Prove
that
for all
nonnegative
real
numbers
x.
Titu
Andreescu,
Gazeta
Matematica
Proof.
This
is
just
a

matter
of
re-arranging
terms
and
applying
Cauchy-
Schwarz:
(ax
2
+ bx +
c)
(cx
2
+ bx +
a)
= (ax
2
+ bx + c)(a + bx + cx
2
)
2 (ax + bx + cx)
2
=
(a+b+c)2x
2
.
D
2.
Let

p
be
a polynomial
with
positive real coefficients.
Prove
that
if
p
(t)
2
p(~)
is
true
for x =
1,
then
it
is
true
for all x >
O.
Titu
Andreescu,
Revista
Matematica
Timi§oara
Proof.
Write
p(X)

=
ao
+
alX
+ +
anxn
and
observe
that
(
1)
n
(al
an)
p(x)p
;:
=(aO+aIX+···+anx)
a
o
+-;:-+···+
xn
2
(ao
+
al
+ + a
n
)2
= p(I)2.
The

result
follows.
D
The
following exercise is
already
a
bit
trickier,
due
to
the
lack of
symmetry.
3.
Prove
that
for all real
numbers
a,
b,
c 2 1
the
following
inequality
holds:
31
Proof.
The
inequality

being
symmetric
in
b,
c,
but
not
in a,
it
is
natural
to
deal
first
with
v1J=1
+ JC=l.
This
is
easy
to
bound
using Cauchy-Schwarz:
v1J=1
+
JC=l
:::;
J(b
- 1 +
1)(1

+ c -
1)
=
V;;;;.
So,
it
is
enough
to
prove
that
VbC
+
va=I
:::;
J a(
bc
+ 1).
But
this
is once
more
the
Cauchy-Schwarz inequality. D
Another
easy,
but
a
bit
exotic

application
of
the
Cauchy-Schwarz inequal-
ity
is
the
following Chinese
olympiad
problem.
4.
Let
n
be
a positive integer.
Find
the
number
of
ordered
n-tuples
of
integers
(a
I,
a2,

, an)
such
that

al
+
a2
+ +
an
2 n
2
and
aI +
a§
+ +
a;
:::;
n
3
+
1.
China
2002
proof
By
the
Cauchy-Schwarz
inequality
we have
al
+
a2
+ +
an

:::;
vn(n3
+
1)
< n
2
+
1.
Since
ai
are integers
and
al
+
a2
+ +
an
2 n
2
,
we
must
have
al
+
a2
+ +
an
= n
2

.
But
then
(again
by
Cauchy-Schwarz) we have
ai
+
a§
+ +
a;
2 n
3
,
forcing
ai
+
a§
+ +
a;
E
{n
3
, n
3
+
I}.
If
ai
+

a~
+ +
a;
= n
3
,
then
we
must
have
equality
in Cauchy-Schwarz,
implying
that
all ai's
are
equal
to
n.
Assume
now
that
ai
+
a§
+ +
a;
= n
3
+ 1

and
let b
i
=
ai
- n.
Then
bI
+
b§
+ +
b~
= n
3
+ 1 -
2n
. n
2
+ n
3
= 1,
forcing all
but
Olle
b
i
vanish.
This
is however impossible,
as

b
l
+b2+
.
-+b
n
=
O.
Therefore,
this
second case will
not
occur
and
the
only
solution
is
D
We
continue
the
series
of
easy
exercises: