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Chapter 3
Homogenizations and Normalizations
Every Mathematician Has Only a Few Tricks. A long time ago an older and well-known number theorist made
some disparaging remarks about Paul Erdös ′ s work. You admire Erdos ′ s contributions to mathematics as much as I
do, and I felt annoyed when the older mathematician flatly and definitively stated that all of Erdos ′ s work could be
reduced to a few tricks which Erdös repeatedly relied on in his proofs. What the number theorist did not realize is
that other mathematicians, even the very best, also rely on a few tricks which they use over and over. Take Hilbert.
The second volume of Hilbert ′ s collected papers contains Hilbert ′ s papers in invariant theory. I have made a point of
reading some of these papers with care. It is sad to note that some of Hilbert ′ s beautiful results have been completely
forgotten. But on reading the proofs of Hilbert ′ s striking and deep theorems in invariant theory, it was surprising to
verify that Hilbert ′ s proofs relied on the same few tricks. Even Hilbert had only a few tricks! Gian-Carlo Rota,
Ten Lessons I Wish I Had Been Taught, Notices of the AMS, January 1997
3.1 Homogenizations
Many inequality problems come with constraints such as ab = 1, xyz = 1, x+y +z = 1. A non-homogeneous
symmetric inequality can be transformed into a homogeneous one. Then we apply two powerful theorems :
Shur’s inequality and Muirhead’s theorem. We begin with a simple example.
Problem 21. (Hungary 1996) Let a and b be positive real numbers with a + b = 1. Prove that
a 2
a + 1 +
b2
b + 1 ≥ 1 3 .
Solution. Using the condition a + b = 1, we can reduce the given inequality to homogeneous one, i. e.,
1
3 ≤ a 2
(a + b)(a + (a + b)) + b 2
(a + b)(b + (a + b)) or a2 b + ab 2 ≤ a 3 + b 3 ,
which follows from (a 3 +b 3 )−(a 2 b+ab 2 ) = (a−b) 2 (a+b) ≥ 0. The equality holds if and only if a = b = 1 2 .
The above inequality a 2 b + ab 2 ≤ a 3 + b 3 can be generalized as following :
Theorem 3.1.1. Let a 1 , a 2 , b 1 , b 2 be positive real numbers such that a 1 + a 2 = b 1 + b 2 and max(a 1 , a 2 ) ≥
max(b 1 , b 2 ). Let x and y be nonnegative real numbers. Then, we have x a1 y a2 + x a2 y a1 ≥ x b1 y b2 + x b2 y b1 .
Proof. Without loss of generality, we can assume that a 1 ≥ a 2 , b 1 ≥ b 2 , a 1 ≥ b 1 . If x or y is zero, then
it clearly holds. So, we assume that both x and y are nonzero. It follows from a 1 + a 2 = b 1 + b 2 that
a 1 − a 2 = (b 1 − a 2 ) + (b 2 − a 2 ). It’s easy to check
x a 1
y a 2
+ x a 2
y a 1
− x b 1
y b 2
− x b 2
y b 1
= x a 2
y a 2 ( x a 1−a 2
+ y a 1−a 2
− x b 1−a 2
y b 2−a 2
− x b 2−a 2
y b 1−a 2
)
= x a2 y ( a2 x b1−a2 − y b1−a2) ( x b2−a2 − y b2−a2)
1 (
=
x a 2 y
a 2
x
b 1
− y b ) ( 1
x b 2
− y b ) 2
≥ 0.
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