Difference between revisions of "2012 USAJMO Problems/Problem 3"
(→Solution 2) |
(→See Also) |
||
Line 64: | Line 64: | ||
{{USAJMO newbox|year=2012|num-b=2|num-a=4}} | {{USAJMO newbox|year=2012|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} |
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.