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1999 USAMO Problems/Problem 4

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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} \qquada_{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 . 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 ... Since k<n, 4(n-k) > 4. It follows that \max(a_1, \dotsc, a_n) \ge \sqrt{4} = 2, as desired. \blacksquare

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