Difference between revisions of "1999 USAMO Problems/Problem 4"
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[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
Revision as of 13:17, 15 September 2012
Problem
Let 
(
) be real numbers such that
![\[a_{1} + a_{2} + \cdots + a_{n} \geq n \qquad \mbox{and} \qquad a_{1}^{2} + a_{2}^{2} + \cdots + a_{n}^{2} \geq n^{2}.\]](http://latex.artofproblemsolving.com/b/4/e/b4e04b54c2d1f0dad88a502d7cd3313552e78b32.png)
Prove that 
.
Solution
First, suppose all the 
are positive. Then
![\[\max(a_1, \dotsc, a_n) \ge \sqrt{\frac{a_1^2 + \dotsb + a_n^2}{n}} \ge \sqrt{n} \ge 2 .\]](http://latex.artofproblemsolving.com/3/5/7/3577ed2f8c34f0c8fad982e812743fd964a1ca09.png)
Suppose, on the other hand, that without loss of generality,
![\[a_1 \ge a_2 \ge \dotsb \ge a_k \ge 0 > a_{k+1} \ge \dotsb \ge a_n,\]](http://latex.artofproblemsolving.com/e/e/0/ee078bb1a2ad809ed0ae9b32af1a26c713de7de6.png)
with 
. If 
we are done, so suppose that 
. Then
, so
![\[\sum_{i=k+1}^n -a_i \le 2k-n .\]](http://latex.artofproblemsolving.com/f/3/e/f3e5202c11b585e83d51b3a687e17520aa1f6255.png)
Since 
is a positive real for all 
, 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
Since 
, 
. It follows that 
, as desired. 
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 | ||
