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:

$$ \frac{n^2}{2(a_1 + a_2 + \dots + a_n)} + \sum_{k=1}^{n} \frac{k}{a_1 + a_2 + \dots + a_k} \le \sum_{j=1}^{n} \frac{2}{a_j}$$
Forgive me as it is my first time on math exchange and I dont know how formatting is done for expressions.

first thing I feel is factor of 2 on right side is not necessary. It can be any number greater than or equal to 1. Equality can occur only when all a_i are identical and that factor on right side is 1.

lets call  expression on right side as RHS and on left side as LHS.

Define f(n) = RHS_n - LHS_n
then f(n+1) = RHS_{n+1} - LHS_{n+1}


it is easy to see that 

f(n+1) - f(n)&gt;0 for all n


(final expression will be quadratic in  a_{n+1} and X, where X = summation of a_i [i =1 to n]). 

given inequality is obvious when n =1, so that f(1) >0. hence f(n) > 0 for all n.