Maths Olympiad Prep

Track / Stage 6 / 48 of 400 #1048 of 1964

Problem 1048

National olympiad, first round
Algebra Difficulty 6.1 Prove it

Example 3 Prove that for any positive numbers a1,a2,⋯ ,an,n⩾2a_{1}, a_{2}, \cdots, a_{n}, n \geqslant 2, we have ∑i=1nais−ai⩾nn−1\sum_{i=1}^{n} \frac{a_{i}}{s-a_{i}} \geqslant \frac{n}{n-1}, where s=∑i=1nais=\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 bi=s−ai>0,i=1,2,⋯ ,nb_{i}=s-a_{i}>0, i=1,2, \cdots, n, then ∑i=1nbi=(n−1)s\sum_{i=1}^{n} b_{i}=(n-1) s, and by the arithmetic mean inequality we get
∑i=1nais−ai=∑i=1ns−bibi=s∑i=1n1bi−n⩾sn1b11b2⋯1bnn−n=sn1b1b2⋯bnn−n⩾sn2b1+b2+⋯+bn−n=sn2(n−1)s−n=nn−1. \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 and solution reproduced as published; topic, difficulty and ordering added by this site.