Difference between revisions of "1994 USAMO Problems/Problem 4"
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Revision as of 10:47, 12 April 2011
Problem 4
Let be a sequence of positive real numbers satisfying for all . Prove that, for all
Solution
Since each is positive, by Muirhead's inequality, . Now we claim that
, giving works, but we set the base case , which gives . Now assume that it works for . By our assumption, now we must prove that, for case, , which is clearly true for . So we are done.
See Also
1994 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |