MathLabs

Problem 5

Let n be an integer at least 2. Prove that if real numbers a_1, a_2, ..., a_n satisfy a_1^2+a_2^2+...+a_n^2=n, then the sum over 1≤i<j≤n of 1/(n-a_i a_j) is at most n/2.
Step 6 of 6: Return to the reciprocal sum
Previous stepNext step
1n−aiaj=1n(1+aiajn−aiaj),∑i<j1n−aiaj≤n−12+12=n2.\frac1{n-a_i a_j}=\frac1n\left(1+\frac{a_i a_j}{n-a_i a_j}\right),\quad \sum_{i<j}\frac1{n-a_i a_j}\le\frac{n-1}{2}+\frac12=\frac n2.
Detailed analysis

Use n/(n-a_i a_j)=1+a_i a_j/(n-a_i a_j). There are n(n-1)/2 pairs, and the preceding bound gives the claimed n/2 upper bound.