Inequality of arithmetic and geometric means

Remarks

For $n = 2$, the inequality $\sqrt{x y} \le (x + y)/2$ follows from $(\sqrt{x} - \sqrt{y})^2 \ge 0$. The weighted form: for weights $w_i \ge 0$ with $w_1 + \dots + w_n = 1$, $x_1^{w_1} \cdots x_n^{w_n} \le w_1 x_1 + \dots + w_n x_n$.