The Art and Craft of Problem Solving

178 CHAPTER 5 ALGEBRA

As the distance between two positive numbers decreases, their product

increases, provided that their sum stays constant.

This agrees with our intuition: As a rectangle becomes more "squarish," i.e., more

symmetrical, it encloses area more "efficiently."

Here is a nice geometric proof of AM-GM. Let AC be the diameter of a circle, and

let B be any point on the circle. Recall that ABC will be a right triangle. Now locate

point D so that BD is perpendicular to AC.

A"--------':--------------�x y C

Then triangles ABD and BCD are similar; hence

x g

g y

Thus g = VXY, the geometric mean of x and y. Indeed, that's why it is called a geo­

metric mean!

Now, let the point B move along the circle, with D moving as well so that BD stays

perpendicular to AC. It is clear that BD is largest when D is at the center of the circle,

in which case x and y are equal (to the length of the radius). Moreover, as D moves

towards the center, x and y become closer in distance, and BD increases.

The AM-GM inequality is true for any finite number of variables. Let

XI,X 2 ,··· ,Xn

be positive real numbers, and define the arithmetic mean An and geometric mean Gn

respectively by

The general version of AM-GM asserts that An 2 Gn, with equality if and only if

XI =X 2 = ... =xn·

There are many ways to prove this (see Problem 5.5.26 on page 186 for hints about

an ingenious induction proof). We shall present a simple argument that uses two strate­

gic ideas: an algorithmic proof style, plus a deliberate appeal to physical intuition. We