Proof that $\sum\limits_{j,k=1}^N\frac{a_ja_k}{j+k}\ge0$ I found this inequality holds:


  Given a sequence of real numbers,
  $a_1,a_2,...,a_N$, if
  $$S=\sum_{j,k=1}^N\frac{a_ja_k}{j+k}$$ then $$S\ge0$$


Is it a known result? How is it possible to prove it?
It strongly reminds me of the Hilbert matrix, although slightly different. And here is a simple and brutal proof:

$$ S
= \sum_{j,k=1}^{N} a_j a_k \int_{0}^{1} x^{j+k-1} \, dx
= \int_{0}^{1} \bigg( \sum_{j=1}^{N} a_j x^{j-1} \bigg)^2 x \, dx
\geq 0 $$

It is also interesting to notice that the equality holds if and only if $a_1 = \cdots = a_N = 0$.
In fact, no two real numbers $a_j,a_k$ will produce a negative result regardless of the numbers, so when it's broadened to $N$ numbers it won't be negative since it's just the sum of multiple sets of 2-variable solutions. $$\frac{a_j^2}{2j}+\frac{a_ja_k}{j+k}+\frac{a_k^2}{2k}$$ The above is the expanded version of a 2-variable solution. This function will never be negative as long as $j,k$ are positive; it's irrelevant how large or small they are.