Track / Stage 7 / 88 of 300 #1488 of 1964

Problem 1488

National olympiad second round; IMO P1/P4

Algebra Difficulty 7.2 Prove it

There are $n \geqslant 3$ positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$ . For each $1 \leqslant i \leqslant n$ we let $b_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}$ (here we define $a_{0}$ to be $a_{n}$ and $a_{n+1}$ to be $a_{1}$ ). Assume that for all $i$ and $j$ in the range 1 to $n$ , we have $a_{i} \leqslant a_{j}$ if and only if $b_{i} \leqslant b_{j}$ .

Prove that $a_{1}=a_{2}=\cdots=a_{n}$ .

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

Suppose that not all $a_{i}$ are equal. Consider an index $i$ such that $a_{i}$ is maximal and $a_{i+1}<a_{i}$ . Then

$b_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}<\frac{2 a_{i}}{a_{i}}=2.$

But since $a_{i}$ is maximal, $b_{i}$ is also maximal, so we must have $b_{j}<2$ for all $j \in\{1,2, \ldots, n\}$ . However, consider the product $b_{1} b_{2} \ldots b_{n}$ . We have

$\begin{aligned} b_{1} b_{2} \ldots b_{n} & =\frac{a_{n}+a_{2}}{a_{1}} \cdot \frac{a_{1}+a_{3}}{a_{2}} \cdot \ldots \cdot \frac{a_{n-1}+a_{1}}{a_{n}} \\ & \geqslant 2^{n} \frac{\sqrt{a_{n} a_{2}} \sqrt{a_{1} a_{3}} \ldots \sqrt{a_{n-1} a_{1}}}{a_{1} a_{2} \ldots a_{n}} \\ & =2^{n} \end{aligned}$

where we used the inequality $x+y \geqslant 2 \sqrt{x y}$ for $x=a_{i-1}, y=a_{i+1}$ for all $i \in\{1,2, \ldots, n\}$ in the second row.

Since the product of all $b_{i}$ is at least $2^{n}$ , at least one of them must be greater than 2, which is a contradiction with the previous conclusion.

Thus, all $a_{i}$ must be equal.

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