Inequality – Arithmetic Mean is Less or Equal Than the Quadratic Mean

Question

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!

Answer 1

By Cauchy–Schwarz inequality, $$\left(\frac{a_1}{n}1+\frac{a_2}{n}1+\cdots+\frac{a_n}{n}1\right)^2\le\left(\frac{a_1^2}{n^2}+\frac{a_2^2}{n^2}+\cdots+\frac{a_n^2}{n^2}\right)\left(1^2+1^2+\cdots+1^2\right)$$