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Revision as of 17:09, 3 July 2013
Contents
[hide]Problem
Let ,
,
be positive real numbers. Prove that
Solution
By the Cauchy-Schwarz inequality,
so
Since
,
Hence,
Again by the Cauchy-Schwarz inequality,
so
Since
,
Hence,
Therefore,
Solution 2
Split up the numerators and multiply both sides of the fraction by either a or 3a:
Then use Titu's Lemma:
It suffices to prove that
After some simplifying, it reduces to
which is trivial by the Rearrangement Inequality.
Solution 3
We proceed to prove that
(then the inequality in question is just the cyclic sum of both sides, since
)
Indeed, by AP-GP, we have
and
Summing up, we have
which is equivalent to:
Dividing from both sides, the desired inequality is proved.
--Lightest 15:31, 7 May 2012 (EDT)
See Also
2012 USAJMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.