Show that the arithmetic mean is less or equal than the quadratic mean I tried to solve this for hours but no success.

Prove that the arithmetic mean is less or equal than the quadratic mean.

I am in front of this form:
$$
\left(\frac{a_1 + ... + a_n} { n}\right)^2 \le \frac{a_1^2 + ... + a_n^2}{n}
$$
With rewriting the inequality in other forms I had no luck.
I think maybe induction would be OK, but I have no idea, how to do it in this case. Do you know a good proof for this?
Thanks!

This is kind of a simplistic and brute-force approach:
$$\bigg(\frac{\sum_{i = 1}^n a_i}{n}\bigg)^2 \leq \frac{\sum_{i=1}^n a_i^2}{n} \iff %
\bigg(\sum_{i = 1}^n a_i\bigg)^2 \leq n \sum_{i=1}^n a_i^2$$
Now open the bracket on the left of the inequality and simplify and note that the result is equivalent to:
$$\sum_{1 \; \leq \; i \; < \; j \; \leq \; n} (a_i - a_j)^2 \geq 0$$
This is obvious.
