The AM-GM inequality

The most basic arithmetic mean-geometric mean (AM-GM) inequality states simply that if x and y are nonnegative real numbers, then $(x+y)/2 \ge \sqrt{xy}$ , with equality if and only if x=y. The last phrase ``with equality...'' means two things: first, if $x=y \ge 0$ , then $(x+y)/2 = \sqrt{xy}$ (obvious); and second, if $(x+y)/2 = \sqrt{xy}$ for some $x,y \ge 0$ , then x=y. It follows that if $x,y \ge 0$ and $x \not= y$ , then inequality is strict: $(x+y)/2 >\sqrt{xy}$ .

Here's a one-line proof of the AM-GM inequality for two variables:

$\begin{displaymath}\frac{x+y}{2}-\sqrt{xy} = \frac{1}{2} \left(\sqrt{x}-\sqrt{y}\right)^2 \ge 0.\end{displaymath}$

The AM-GM inequality generalizes to n nonnegative numbers.

AM-GM inequality:
If $x_1,\dots,x_n \ge 0$ , then

$\begin{displaymath}\frac{x_1+x_2+\dots+x_n}{n} \ge \sqrt[n]{x_1 x_2 \dots x_n}\end{displaymath}$

with equality if and only if $x_1=x_2=\dots=x_n$ .