# Arithmetic mean always greater than geometric mean

Hey,

(sin A + sin B + sin C)/3 >= $$\sqrt[3]{}$$(sin A*sin B*sin C)

I know this is true by Arithmetic mean always greater than geometric mean...
but is there any other way of proving this?

$$\left[ \frac13 (\sin A + \sin B + \sin C) \right]^3 \ge \frac19 \left( \sin^3 A + \sin^3 B + \sin^3 C \right) \ge \frac13 \sin^3 A \ge \sin A \sin B \sin C$$