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Prove that $\sum_{i=1}^n \sum_{j=1}^n \dfrac{a_i a_j}{i+j-1} \ge 0$ for any real numbers $a_1,\cdots, a_n$.

One possible approach might be to find the minimum possible value of $\sum_{i=1}^n \sum_{j=1}^n \dfrac{a_i a_j}{i+j-1}$ or at least find a lower bound for it that's at least 0. It would be useful if we could reduce the number of possibilities we'd need to consider for the $a_i$'s (e.g. for linear functions, we can just maximize or minimize by checking boundary points, but here we don't have any boundary points). The expression is a continuous function of $a_1,\cdots, a_n,$ so it attains a maximum and minimum on any compact subset of $\mathbb{R}^n.$ It might be useful to consider the effect of swapping any two $a_i$'s. Maybe the inequality becomes easier to prove if we assume $a_1\leq a_2\leq \cdots \leq a_n$?

Edit: in hindsight, this is equivalent to showing a Hilbert matrix is positive definite. The main reason why I couldn't solve this question initially is because I didn't realize the fact that the given expression can be rewritten as $\sum_{i=1}^n \sum_{j=1}^n a_i a_j\int_0^1 t^{i-1+j-1} dt = \int_0^1 (\sum_{i=1}^n a_i t^{i-1})^2 dt.$ This notion of writing things as sums of squares is actually a very useful trick for proving inequalities and can be used for math contests.

user3379
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    If I remember correctly, the expression in your OP can be expressed as $\int^1_0 P^2_n(x),dx$ where $P_n(x)=a_1+a_2x+\ldots +a_nx^{n-1}$. – Mittens Oct 01 '22 at 17:11

1 Answers1

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No need to make special assumptions. It is always true.

Write your condition under the form

$$X^THX \ge 0 \ \ \text{for} \ X=(a_1,...a_n)^T$$

where $H$ is a Hilbert matrix, which is known to be positive definite.

In fact the result is $>0$ unless all the $a_k$ are zero.

Jean Marie
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  • I am not the downvoter 2) There are people (certainly few) on this site that are compulsive downvoters and are harmful. I have been many times the "victim" of them. 3) You have indeed visibly worked on your issue ; for next time , think to add supplementary information : the fact that it is either a homework (and in this case which discipline : linear algebra ? or analysis ? or optimization ? In certain case, even the chapter you are studying : quadratic forms ? ) or, who knows, some kind of preparation for some Olympiads... 4) Would you mind explaining why my answer hasn't been upvoted ?
    • – Jean Marie Oct 02 '22 at 09:28
  • Thanks. Upvoted. – user3379 Oct 02 '22 at 18:14