AlgebraAMC 10

AM-GM Inequality

\frac{a+b}{2} \ge \sqrt{ab}

For nonnegative reals $a, b$ ; equality when $a = b$ .

What it says

The arithmetic mean of nonnegative numbers is always at least their geometric mean, with equality exactly when the numbers are equal. It turns "find the minimum" questions into one line: if a product is fixed, the sum is smallest when the terms are equal. It extends to any number of terms.

Show proof

Since squares are nonnegative, $(\sqrt a - \sqrt b)^2 \ge 0$ . Expanding gives $a - 2\sqrt{ab} + b \ge 0$ , so $\frac{a + b}{2} \ge \sqrt{ab}$ , with equality only when $\sqrt a = \sqrt b$ .

When to use it

You need the minimum of a sum whose product is fixed, or the maximum of a product whose sum is fixed.

Example

Find the minimum value of $x + \frac{9}{x}$ for $x > 0$ .

Show solution

$x + \frac9x \ge 2\sqrt{9} = 6$ , with equality at $x = 3$ . Answer: $6$