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Revision as of 11:55, 17 September 2012

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.

See Also

1999 USAMO (Problems • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions
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