The Art and Craft of Problem Solving

5.5 INEQUALITIES 177

Consider the following equivalent formulation. Let x,y be positive real numbers

with sum S = x + y. Then the maximum value of the product xy is attained when x and

yare equal, i.e., when x = y = S /2. In other words, we assert that

(�) 2

=

(x:y)^2

2 xy.

It is a simple matter to prove this. We will place a question mark over the ":S;" to
remind ourselves that it is an "alleged" inequality until we reduce it to an equivalent
one that we know is true. The algebra is simple stuff that we have seen before:

(x+ y)^2?

4

2xy

is equivalent to

?

(x+y)^2 � (^4) xy.
Thus
?
x^2 + 2xy + i � (^4) xy.
Subtracting (^4) xy from both sides yields
?
x^2 -2xy+i � 0,

and now we can remove the question mark, for the left-hand side is the square (x ­

y)^2 , hence always non-negative. We have proven the inequality, and moreover, this

argument also shows that equality occurs only when x = y, for only then will the

square equal zero.
This inequality is called the Arithmetic-Geometric Mean Inequality, often ab­
breviated as AM-GM or AGM, and is usually written in the form

X

;

y

2 y'XY.

Recall that the left-hand side is the arithmetic mean of x and y, while the right-hand

side is called the geometric mean. A succinct way to remember this inequality is that

The arithmetic mean of two positive real numbers is greater than or

equal to the geometric mean, with equality only if the numbers are

equal.

We can extract more information from our algebraic proof of AM-GM. Since (x+

y)^2 2 4 xy and (x+y)^2 -4xy = (x - y)^2 , we can write

S^2 -4P =D^2 ,

where S, P, D are respectively the sum, product, and difference of x and y. If we let x

and y vary so that their sum S is fixed, we see that the product of x and y is a strictly

decreasing function of the distance between x and y. (By distance, we mean Ix - yl.)

This is so useful, we shall name it the symmetry-product principle: