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 5 of 6: Sum over all pairs
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∑i<jaiajn−aiaj≤12∑i≠jai2n−aj2=12∑j=1nn−aj2n−aj2=n2.\sum_{i<j}\frac{a_i a_j}{n-a_i a_j}\le\frac12\sum_{i\ne j}\frac{a_i^2}{n-a_j^2}=\frac12\sum_{j=1}^n\frac{n-a_j^2}{n-a_j^2}=\frac n2.
Detailed analysis

Sum the pair estimates. Reindexing the resulting double sum by the denominator index j, the numerator sum over i≠j is n-a_j^2. The result is n/2.