Fall 2018, problem 74

Prove that, for any real numbers $a_1,a_2, \ldots, a_n$, $\displaystyle\qquad\sum_{i,j=1}^n{\frac{a_ia_j}{i+j-1}}\geq 0$.


3 years ago

$0\leq\int_0^1\big(\sum_{i=1}^na_ix^{i-1}\big)^2dx=\int_0^1 \big(\sum_{i=1}^na_ix^{i-1}.\sum_{j=1}^na_ix^{j-1}\big)dx$

$=\int_0^1 \big(\sum_{i,j=1}^na_ia_jx^{i+i-2}dx\big)=\sum_{i,j=1}^n\frac {a_ia_j}{i+j-1}.$

  • with equality for all zero -
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: $$\sum_{i,j=1}^n \frac{a_ia_j}{b_i+b_j-1}\ge0$$

3 years ago

It's equivalent to prove Hilbert matrix is positive definite.