Difference between revisions of "1999 USAMO Problems/Problem 4"
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<cmath> \sum_{i=k+1}^n -a_i \le 2k-n . </cmath> | <cmath> \sum_{i=k+1}^n -a_i \le 2k-n . </cmath> | ||
Since <math>-a_i</math> is a positive real for all <math>k+1 \le i \le n</math>, it follows that | Since <math>-a_i</math> is a positive real for all <math>k+1 \le i \le n</math>, it follows that | ||
− | <cmath> \sum_{i=k+1}^n a_i^2 \le \left( \sum_{i=k+1}^n | + | <cmath> \sum_{i=k+1}^n a_i^2 \le \left( \sum_{i=k+1}^n-a_i \right)^2 \le (2k-n)^2 . </cmath> |
Then | Then | ||
<cmath> \begin{align*} | <cmath> \begin{align*} |
Revision as of 09:37, 2 July 2015
Contents
Problem
Let () be real numbers such that Prove that .
Hint
Remember that the problem said real numbers, which can be negative!
Solution
First, suppose all the are positive. Then Suppose, on the other hand, that without loss of generality, with . If we are done, so suppose that . Then , so Since is a positive real for all , it follows that Then Since , . It follows that , as desired.
Solution 2
Assume the contrary and suppose each is less than 2.
Without loss of generality let , and let be the largest integer such that and if it exists, or 0 if all the are non-negative. If , then (as ) , a contradiction. Hence, assume . Then Because for , both sides of the inequality are non-positive, so squaring flips the sign. But we also know that for , so which results in a contradiction to our given condition. The proof is complete.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.