$(\sum_{k=1}^{n}a_{k}x^{k-1})^2=\sum_{i,j=1}^{n}a_{i}a_{j}x^{i+j-2}$. $0\le \int_{0}^{1}\sum_{i,j=1}^{n}a_{i}a_{j}x^{i+j-2}dx$ $=\sum_{i,j=1}^{n}\int_{0}^{1}a_{i}a_{j}x^{i+j-2}dx$ $=\sum_{i,j=1}^{n}\frac{a_{i}a_{j}}{i+j-1}$ Q.E.D.

djb86 wrote: Prove that, for any real numbers $a_1, a_2, \dots , a_n$, \[ \sum_{i,j=1}^n \frac{a_ia_j}{i+j-1}\ge 0.\] The following is also true: Prove that, for any real numbers $a_1, a_2, \dots , a_n$, \[ \sum_{i,j=1}^n \frac{a_ia_j}{i+j}\ge 0.\]proof is similar to above.

The equality $\sum_{i,j=1}^n \frac{a_ia_j}{i+j-1}= 0$ comes iff $a_1=a_2=...=a_n=0$

Amazing topic