AM–GM Inequality (Two Variables)

Statement

For $a,b\ge 0$, \[ \frac{a+b}{2} \ge \sqrt{ab}, \] with equality iff $a=b$.

Proof

Since $(\sqrt a-\sqrt b)^2\ge 0$, \[ a - 2\sqrt{ab} + b \ge 0 \ \Rightarrow\ a+b \ge 2\sqrt{ab} \ \Rightarrow\ \frac{a+b}{2} \ge \sqrt{ab}. \] Equality $\Leftrightarrow (\sqrt a-\sqrt b)^2=0 \Leftrightarrow a=b$.