Difference between revisions of "1997 USAMO Problems/Problem 5"
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== Solution 2 == | == Solution 2 == | ||
− | Rearranging the AM-HM inequality, we get <math>\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \le \frac{9}{x+y+z}</math>. | + | Rearranging the AM-HM inequality, we get <math>\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \le \frac{9}{x+y+z}</math>. Work in progress. |
==See Also == | ==See Also == |
Revision as of 23:03, 21 August 2020
Contents
Problem
Prove that, for all positive real numbers
.
Solution 1
Because the inequality is homogenous (i.e. can be replaced with
without changing the inequality other than by a factor of
for some
), without loss of generality, let
.
Lemma:
Proof: Rearranging gives
, which is a simple consequence of
and
Thus, by :
Solution 2
Rearranging the AM-HM inequality, we get . Work in progress.
See Also
1997 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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.