# 1998 IMO Shortlist Problems/A1

## Problem

(Mongolia) Let $a_1, a_2, \dotsc, a_n$ be positive real numbers such that $a_1 + a_2 + \dotsb + a_n < 1$. Prove that $$\frac{a_1 a_2 \dotsm a_n \bigl[ 1- (a_1 + a_2 + \dotsb + a_n) \bigr]}{ (a_1 + a_2 + \dotsb + a_n)(1-a_1)(1-a_2) \dotsm (1-a_n)} \le \frac{1}{n^{n+1}} .$$

## Solution

Let $a_{n+1}$ be the positive real number such that $a_{n+1} = 1 - \sum_{k=1}^n a_k$. Then $$\frac{a_1 a_2 \dotsm a_n \bigl[ 1- (a_1 + a_2 + \dotsb + a_n) \bigr]}{ (a_1 + a_2 + \dotsb + a_n)(1-a_1)(1-a_2) \dotsm (1-a_n)} = \frac{\prod_{k=1}^{n+1} a_k} {\prod_{k=1}^{n+1} \sum_{j\neq k} a_j} .$$ But by AM-GM, $\sum_{j\neq k} a_j \ge n \prod_{j\neq k} a_j^{1/n}$. Hence $$\frac{\prod_{k=1}^{n+1} a_k}{\prod_{k=1}^{n+1} \sum_{j\neq k} a_j} \le \frac{\prod_{k=1}^{n+1} a_k}{\prod_{k=1}^{n+1} n \prod_{j\neq k} a_j^{1/n}} = \frac{\prod_{k=1}^{n+1} a_k}{n^{n+1} \prod_{k=1}^{n+1} a_j} = \frac{1}{n^{n+1}},$$ as desired. Equality holds when all the $a_k$ are equal to $1/(n+1)$. $\blacksquare$