1999 USAMO Problems/Problem 4

Problem

Let $a_{1}, a_{2}, \dots, a_{n}$ ( $n > 3$ ) be real numbers such that $a_{1} + a_{2} + \cdots + a_{n} \geq n \qquad \mbox{and} \qquad a_{1}^{2} + a_{2}^{2} + \cdots + a_{n}^{2} \geq n^{2}.$ Prove that $\max(a_{1}, a_{2}, \dots, a_{n}) \geq 2$ .

Solution

First, suppose all the $a_i$ are positive. Then $\max(a_1, \dotsc, a_n) \ge \sqrt{\frac{a_1^2 + \dotsb + a_n^2}{n}} \ge \sqrt{n} \ge 2 .$ Suppose, on the other hand, that without loss of generality, $a_1 \ge a_2 \ge \dotsb \ge a_k \ge 0 > a_{k+1} \ge \dotsb \ge a_n,$ with $1\le k <n$ . If $a_1 >2$ we are done, so suppose that $a_1 \le2$ . Then $\sum_{i=1}^k a_i \le 2k$ , so $\sum_{i=k+1}^n -a_i \le 2k-n .$ Since $-a_i$ is a positive real for all $k+1 \le i \le n$ , it follows that

\[\sum_{i=k+1}^n a_i^2 \le \left( \sum_{i=k+1}^n} -a_i \right)^2 \le (2k-n)^2 .\] (Error compiling LaTeX. Unknown error_msg)

Then $\begin{align*} \max(a_1, \dotsc, a_n)^2 &\ge \sum_{i=1}^k a_i^2 /k \\ &\ge \left( n^2 - \sum_{i=k+1}^n a_i^2 \right)/k \\ &\ge \frac{n^2 - (2k-n)^2}{k} = 4(n-k). \end{align*}$ Since $k<n$ , $4(n-k) > 4$ . It follows that $\max(a_1, \dotsc, a_n) \ge \sqrt{4} = 2$ , as desired. $\blacksquare$

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1999 USAMO (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
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1999 USAMO Problems/Problem 4

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