A version of Hardy's inequality involving reciprocals.

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:

$$ \frac{n^2}{2(a_1 + a_2 + \dots + a_n)} + \sum_{k=1}^{n} \frac{k}{a_1 + a_2 + \dots + a_k} \le \sum_{j=1}^{n} \frac{2}{a_j}$$

There is also the following N.Sedrakyan's proof.

By C-S $$2\sum_{k=1}^n\frac{1}{a_k}>2\sum_{k=1}^n\frac{k^2}{a_k}\left(\frac{1}{k^2}-\frac{1}{(n+1)^2}\right)=2\sum_{k=1}^n\frac{k^2}{a_k}\sum_{m=k}^n\left(\frac{1}{m^2}-\frac{1}{(m+1)^2}\right)=$$ $$=2\sum_{k=1}^n\left(\frac{1}{k^2}-\frac{1}{(k+1)^2}\right)\sum_{m=1}^k\frac{m^2}{a_m}=2\sum_{k=1}^n\left(\frac{2k+1}{k^2(k+1)^2\sum\limits_{m=1}^ka_m}\sum\limits_{m=1}^ka_m\sum_{m=1}^k\frac{m^2}{a_m}\right)\geq$$ $$\geq2\sum_{k=1}^n\left(\frac{2k+1}{k^2(k+1)^2\sum\limits_{m=1}^ka_m}\left(\sum\limits_{m=1}^km\right)^2\right)=2\sum_{k=1}^n\frac{(2k+1)k^2(k+1)^2}{4k^2(k+1)^2\sum\limits_{m=1}^ka_m}>\sum_{k=1}^n\frac{k}{\sum\limits_{m=1}^ka_m}.$$