1999 USAMO Problems/Problem 4
Problem
Let () be real numbers such that Prove that .
Solution
Solution 1
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 |
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