Fall 2018, problem 74
Prove that, for any real numbers $a_1,a_2, \ldots, a_n$, $\displaystyle\qquad\sum_{i,j=1}^n{\frac{a_ia_j}{i+j-1}}\geq 0$.
Comments
Claudio
3 years ago
Hi, Hubert. What a magnificent solution!
After giving you a "+", I looked in vain for other demonstrations.
The only thing I found is a possible generalization of the result.
With the same demonstration we have:
For any real numbers $a_1,a_2,…,a_n$ and any real exponents $b_1,b_2,...,b_n$ with $b_j>\frac12$ it is: $$\sum_{i,j=1}^n \frac{a_ia_j}{b_i+b_j-1}\ge0$$
Sia
3 years ago
It's equivalent to prove Hilbert matrix is positive definite.
$0\leq\int_0^1\big(\sum_{i=1}^na_ix^{i-1}\big)^2dx=\int_0^1 \big(\sum_{i=1}^na_ix^{i-1}.\sum_{j=1}^na_ix^{j-1}\big)dx$
$=\int_0^1 \big(\sum_{i,j=1}^na_ia_jx^{i+i-2}dx\big)=\sum_{i,j=1}^n\frac {a_ia_j}{i+j-1}.$