Track / Stage 7 / 260 of 300 #1660 of 1964

Problem 1660

National olympiad second round; IMO P1/P4

Algebra Difficulty 7.7 Prove it

$\square$ Example 3 Let $a_{1}, a_{2}, \cdots$ be a sequence of positive real numbers, and for all $i, j=1,2, \cdots$ , satisfy $a_{i+j} \leqslant a_{i}+a_{j}$ . Prove that for any positive integer $n$ , we have $a_{1}+\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{n}}{n} \geqslant a_{n}$ . (1999 Asia Pacific Mathematical Olympiad Problem)

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

Let $S_{i}=a_{1}+a_{2}+\cdots+a_{i}, i=1,2, \cdots, n$ , and define $S_{0}=0$ , then
$2 S_{i}=\left(a_{1}+a_{i}\right)+\left(a_{2}+a_{i-1}\right)+\cdots+\left(a_{i}+a_{1}\right) \geqslant i a_{i+1},$

i.e., $S_{i} \geqslant \frac{i}{2} a_{i+1}$ . Therefore,
$\begin{aligned} & a_{1}+\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{n}}{n} \\ = & \sum_{i=1}^{n} \frac{a_{i}}{i}=\sum_{i=1}^{n} \frac{S_{i}-S_{i-1}}{i} \\ = & \sum_{i=1}^{n-1} S_{i}\left(\frac{1}{i}-\frac{1}{i+1}\right)+\frac{1}{n} S_{n} \\ \geqslant & \frac{1}{2} S_{1}+\sum_{i=1}^{n-1} \frac{i a_{i+1}}{2}\left(\frac{1}{i}-\frac{1}{i+1}\right)+\frac{1}{n} S_{n} \\ = & \frac{1}{2} S_{1}+\frac{1}{2} \sum_{i=1}^{n-1} \frac{a_{i+1}}{i+1}+\frac{1}{n} S_{n}=\frac{1}{2} \sum_{i=1}^{n} \frac{a_{i}}{i}+\frac{1}{n} S_{n} \end{aligned}$

Thus, $\sum_{i=1}^{n} \frac{a_{i}}{i} \geqslant \frac{2}{n} S_{n}=\frac{2}{n}\left(S_{n-1}+a_{n}\right) \geqslant \frac{2}{n}\left(\frac{n-1}{2} a_{n}+a_{n}\right)$
$=\frac{n+1}{n} a_{n}>a_{n} .$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.