TITLE: Is this sum $S=\sum_{i,j=1}^n\frac{a_ia_j}{i+j}$ greater than or equal to zero?
QUESTION [5 upvotes]: Given $$(a_1,a_2,...,a_n)\in\mathbb{R}$$
does this inequality hold?
$$S=\sum_{i,j=1}^n\frac{a_ia_j}{i+j}\ge0$$
Thanks.

REPLY [2 votes]: Let $S(x)=\sum_{i,j} a_i a_j x^{i+j}/(i+j)$. Then $S'(x) =\sum_{i,j} a_i a_j x^{i+j-1}$. We can write $S'(x)=x(\sum_i a_i x^{i-1})^2$ which is nonnegative for nonnegative $x$. Therefore, $S(x)$ is increasing over nonnegative real numbers. In particular, $0=S(0)\le S(1)=S$.