Show that for $a_i>0$ $\frac{a_1+\cdots+a_n}{n}$ converges to $0$ if and only if $\frac{a_1^2+\cdots+a_n^2}{n}$ converges to $0$. Let $\{a_n\}$ be a bounded and positive sequence. Show that
$$\lim_{n\to \infty}\frac{a_1+\cdots+a_n}{n}=0$$
if and only if
$$\lim_{n\to \infty}\frac{a_1^2+\cdots+a_n^2}{n}=0.$$ 
My attempt:
The "$\Rightarrow$" is obvious. Note that 
$$\frac{a_1^2+\cdots+a_n^2}{n}\leq |M|\cdot\frac{a_1+\cdots+a_n}{n} $$
where $|M|$ is the bound of the sequence. So the convergence of the right side implies the convergence of the left side.
As for the converse direction, I really have no idea...

@kimchi lover points out using the Cauchy-Schwarz inequality and I had the following attempt...
$$\frac{a_1+\cdots+a_n}{n}=\frac{\frac{1}{\sqrt{n}}(a_1+\cdots+a_n)}{\frac{1}{\sqrt{n}}n}\leq \frac{(a_1^2+\cdots+a_n^2)(\frac{1}{n}+\cdots+\frac{1}{n})}{\sqrt{n}}$$

By Cauchy—Schwarz inequality,
$$
\sum_{k=1}^n \frac{1}{n}\cdot a_k \leq \sqrt{\sum_{k=1}^n a_k^2}\cdot \sqrt{\sum_{k=1}^n \frac{1}{n^2}}
= \sqrt{\sum_{k=1}^n a_k^2}\cdot\sqrt{\frac{1}{n}}
= \sqrt{\frac{1}{n}\sum_{k=1}^n a_k^2}
$$
and you can conclude by the squeeze theorem.
