Fall 2018, problem 74
Prove that, for any real numbers a1,a2,…,an, n∑i,j=1aiaji+j−1≥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 a1,a2,…,an and any real exponents b1,b2,...,bn with bj>12 it is: n∑i,j=1aiajbi+bj−1≥0
Sia
3 years ago
It's equivalent to prove Hilbert matrix is positive definite.
0≤∫10(∑ni=1aixi−1)2dx=∫10(∑ni=1aixi−1.∑nj=1aixj−1)dx
=∫10(∑ni,j=1aiajxi+i−2dx)=∑ni,j=1aiaji+j−1.