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

Here is a proof.
We try using induction, but as usual, a direct approach seems to fail and we have to try and prove a stronger statement.
So we try and pick a positive function $f(n)$ such that
$$ \frac{f(n)}{a_1 + a_2 + \dots + a_n} + \sum_{k=1}^{n} \frac{k}{a_1 + a_2 + \dots + a_k} \le \sum_{j=1}^{n} \frac{2}{a_j}$$
Let $S = a_1 + a_2 + \dots + a_n$ and let $x = a_{n+1}$.
In order to prove that $n$ implies $n+1$ it would be sufficient to prove
$$\frac{2}{x} + \frac{f(n)}{S} \ge \frac{f(n+1) + n+1}{S+x}$$
This can be rearranged to
$$2S^2 + f(n) x^2 + (f(n) + 2) Sx \ge (f(n+1) + n+1) Sx$$
Since $$ 2S^2 + f(n) x^2 \ge 2 \sqrt{2 f(n)} Sx$$
it is sufficient to prove that $f(n)$ satisfies
$$ f(n) + 2 + 2\sqrt{2 f(n)} \ge f(n+1) + n + 1$$
Choosing $f(n) = \dfrac{n^2}{2}$ does the trick.
We can easily verify the base case for this choice of $f(n)$.
Thus we have:
$$ \frac{n^2}{2(a_1 + a_2 + \dots + a_n)} + \sum_{k=1}^{n} \frac{k}{a_1 + a_2 + \dots + a_k} \le \sum_{j=1}^{n} \frac{2}{a_j}$$
