Geometric average less or equal to arithmetic one

22 Views Asked by At

Let $\ $ ${y_1\cdot...\cdot y_n} \in \mathbb{R}$ $\ $ be positive $\quad$

Prove: $\sqrt[n]{y_1\cdot...\cdot y_n}$ $\le$ $\frac{y_1+...+y_n}{n}$

I have tried to find this by searching keywords like gemometric and arithmetic average, but hadn't found.

Actually, I have no idea how to start proving this one.