# Need help with continuing an idea concerning showing that $4\sum\limits_{n \ge 1} a_n^2 \ge \sum\limits_{n \ge 1} \frac1{n^2}(a_1+…+a_n)^2$

I recently encountered the following problem: If $\sum a_n^2$ converges and $\alpha_n= \frac{a_1+...+a_n}{n}$ then show that:

$$4\sum_{n \ge 1} a_n^2 \ge \sum_{n \ge 1} \alpha_n^2$$

I had an approach which i did not manage to utilize, so i am asking if anyone has an idea if it can be used or not:

1. We note that the right hand side will be maximized if $a_i$ are all of the same sign. Without Loss of Generality, suppose that $a_i \ge 0$.
2. If we use (1.), another way to maximize the right hand side would be if $a_i \ge a_{j}$ for all $i>j$. Let us also suppose that this is true.

And i stop there, i tried many approaches but nothing seemed to work. I thank you in advance for any hint and/or a complete solution.

Supplementary question: Can we construct a similar problem for integrals (instead of series)?

## 1 Answer

This is Hardy's inequality

$$\sum_{n=1}^{\infty}\left(\dfrac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)^p<\left(\dfrac{p}{p-1}\right)^p\sum_{n=1}^{\infty}a^p_{n}$$ take $p=2$