A version of Hardy's inequality involving reciprocals. How can one prove for any sequence of positive numbers $a_n, n\ge1,$ we have
$$\sum_{n=1}^\infty \frac{n}{a_1+a_2+a_3+\cdots+a_n}\le 2\sum_{n=1}^\infty \frac{1}{a_n}$$

Added later:
Apparently, this is a version of Hardy's inequality. The above is the case $p=-1$. (See the wiki for what $p$ is).
The case $p=2$ appears here: Proving $A: l_2 \to l_2$ is a bounded operator

There is also the following N.Sedrakyan's proof.
By C-S
$$2\sum_{k=1}^n\frac{1}{a_k}>2\sum_{k=1}^n\frac{k^2}{a_k}\left(\frac{1}{k^2}-\frac{1}{(n+1)^2}\right)=2\sum_{k=1}^n\frac{k^2}{a_k}\sum_{m=k}^n\left(\frac{1}{m^2}-\frac{1}{(m+1)^2}\right)=$$
$$=2\sum_{k=1}^n\left(\frac{1}{k^2}-\frac{1}{(k+1)^2}\right)\sum_{m=1}^k\frac{m^2}{a_m}=2\sum_{k=1}^n\left(\frac{2k+1}{k^2(k+1)^2\sum\limits_{m=1}^ka_m}\sum\limits_{m=1}^ka_m\sum_{m=1}^k\frac{m^2}{a_m}\right)\geq$$
$$\geq2\sum_{k=1}^n\left(\frac{2k+1}{k^2(k+1)^2\sum\limits_{m=1}^ka_m}\left(\sum\limits_{m=1}^km\right)^2\right)=2\sum_{k=1}^n\frac{(2k+1)k^2(k+1)^2}{4k^2(k+1)^2\sum\limits_{m=1}^ka_m}>\sum_{k=1}^n\frac{k}{\sum\limits_{m=1}^ka_m}.$$
