How can I prove $\sum_{i=1}^n \sum_{j=1}^n \frac{a_ia_j}{i+j-1}$ is never negative for any set of n real numbers $a_i$

Question

I can't figure this out, can someone help me prove it? I know you guys will come up with an incredibly elegant solution

What have you tried? Where are you stuck? Have you tried the $n=2, 3 $ cases? — Calvin Lin, Nov 20 '20 at 23:58
Yes, I tried some particular cases, what gives me trouble is the fact that I can't really know anything about the numbers, — Mr.Konn, Nov 21 '20 at 00:09
This has been asked and answered before, e.g. here: https://math.stackexchange.com/q/1754729/42969. — Martin R, Nov 21 '20 at 10:05
Does this answer your question? Prove the positive definiteness of Hilbert matrix — Parcly Taxel, Nov 22 '20 at 07:40

score 2 · Answer 1 · answered Nov 20 '20 at 23:58

2

Hint: Since $\dfrac{1}{i+j-1} = \displaystyle\int_{0}^{1}x^{i+j-2}\,dx$, we have $$\sum_{i = 1}^{n}\sum_{j = 1}^{n}\dfrac{a_ia_j}{i+j-1} = \sum_{i = 1}^{n}\sum_{j = 1}^{n}\int_{0}^{1}a_ia_jx^{i+j-2}\,dx = \int_{0}^{1}\sum_{i = 1}^{n}a_ix^{i-1} \cdot \sum_{j = 1}^{n}a_jx^{j-1}\,dx$$

Can you see why this is non-negative?

answered Nov 20 '20 at 23:58

JimmyK4542

55,969

(The limits of summation are slightly off. I believe that's on OP rather than you.) – Calvin Lin Nov 20 '20 at 23:59
Hi, thank you for your answer, I got it but I don't get why the last thing is non-negative. – Mr.Konn Nov 21 '20 at 00:07
Because inside the integral the two sums are identical. – Neat Math Nov 21 '20 at 03:26

How can I prove $\sum_{i=1}^n \sum_{j=1}^n \frac{a_ia_j}{i+j-1}$ is never negative for any set of n real numbers $a_i$

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