Track / Stage 6 / 86 of 400 #1086 of 1964

Problem 1086

National Olympiad, first round

Algebra Difficulty 6.1 Prove it

Prove that for any positive numbers $a_{1}, a_{2}, \cdots, a_{n}, n \geqslant 2$ , we have $\sum_{i=1}^{n} \frac{a_{i}}{s-a_{i}} \geqslant \frac{n}{n-1}$ , where $s=\sum_{i=1}^{n} a_{i}$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Prove that if $b_{i}=s-a_{i}>0, i=1,2, \cdots, n$ , then $\sum_{i=1}^{n} b_{i}=(n-1) s$ , and by the arithmetic mean inequality we get
$\begin{array}{l} \sum_{i=1}^{n} \frac{a_{i}}{s-a_{i}}=\sum_{i=1}^{n} \frac{s-b_{i}}{b_{i}}=s \sum_{i=1}^{n} \frac{1}{b_{i}}-n \geqslant s n \sqrt[n]{\frac{1}{b_{1}} \frac{1}{b_{2}} \cdots \frac{1}{b_{n}}}-n \\ =s n \frac{1}{\sqrt[n]{b_{1} b_{2} \cdots b_{n}}}-n \geqslant \frac{s n^{2}}{b_{1}+b_{2}+\cdots+b_{n}}-n=\frac{s n^{2}}{(n-1) s}-n=\frac{n}{n-1} . \end{array}$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.