# 2009 USAMO Problems/Problem 4

## Problem

For $n \ge 2$ let $a_1$, $a_2$, ..., $a_n$ be positive real numbers such that

$(a_1+a_2+ ... +a_n)\left( {1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n} \right) \le \left(n+ {1 \over 2} \right) ^2$

Prove that $\text{max}(a_1, a_2, ... ,a_n) \le 4 \text{min}(a_1, a_2, ... , a_n)$.

## Solution

Assume without loss of generality that $a_1 \geq a_2 \geq \cdots \geq a_n$. Now we seek to prove that $a_1 \le 4a_n$.

By the Cauchy-Schwarz Inequality, \begin{align*} (a_n+a_2+ a_3 + ... +a_{n-1}+a_1)\left({1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n}\right) &\ge \left( \sqrt{a_n \over a_1} + n-2 + \sqrt{a_1 \over a_n} \right)^2 \\ (n+ {1 \over 2})^2 &\ge \left( \sqrt{a_n \over a_1} + n-2 + \sqrt{a_1 \over a_n} \right)^2 \\ n+ {1 \over 2} &\ge n-2 + \sqrt{a_n \over a_1} + \sqrt{a_1 \over a_n} \\ {5 \over 2} &\ge \sqrt{a_n \over a_1} + \sqrt{a_1 \over a_n} \\ {17 \over 4} &\ge {a_n \over a_1} + {a_1 \over a_n} \\ 0 &\ge (a_1 - 4a_n)\left(a_1 - {a_n \over 4}\right) \end{align*} Since $a_1 \ge a_n$, clearly $(a_1 - {a_n \over 4}) > 0$, dividing yields:

$$0 \ge (a_1 - 4a_n) \Longrightarrow 4a_n \ge a_1$$

as desired.

Alternative Solution (by Deng Tianle, username: Leole) Assume without loss of generality that $a_1 \geq a_2 \geq \cdots \geq a_n$. Using the Cauchy–Bunyakovsky–Schwarz inequality and the inequality given, \begin{align*} &\left(a_1+a_2 + ... +a_n + 3a_n -\frac{3a_1}{4}\right)\left({1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n}\right) \\ =&\left(\frac{a_1}{4}+a_2 + ... +a_{n-1}+4a_n\right)\left({1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n}\right) \\ \ge& \left(\frac{1}{2}+n-2+2\right)^2 \\ =&\left(n+\frac{1}{2}\right)^2 \\ \ge& (a_1+a_2 + ... +a_{n})\left({1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n}\right).\end{align*} (Note that $n-2 \ge 0$ since $n \ge 2$ as given!) This implies that $3a_n -\frac{3a_1}{4} \ge 0 \iff 4a_n \ge a_1$ as desired.