Difference between revisions of "2012 USAJMO Problems/Problem 3"
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Then use Titu's Lemma: <math>\sum_{cyc} \frac {a^4} {5a^2+ab} +\sum_{cyc} \frac {9a^4} {15ac+3a^2} \ge \frac {16(a^2+b^2+c^2)^2} {8(a^2+b^2+c^2)+16(ab+bc+ca)}</math> | Then use Titu's Lemma: <math>\sum_{cyc} \frac {a^4} {5a^2+ab} +\sum_{cyc} \frac {9a^4} {15ac+3a^2} \ge \frac {16(a^2+b^2+c^2)^2} {8(a^2+b^2+c^2)+16(ab+bc+ca)}</math> | ||
It suffices to prove that<math> \frac {16(a^2+b^2+c^2)^2} {8(a^2+b^2+c^2)+16(ab+bc+ca)} \ge \frac {2} {3} (a^2+b^2+c^2)</math> | It suffices to prove that<math> \frac {16(a^2+b^2+c^2)^2} {8(a^2+b^2+c^2)+16(ab+bc+ca)} \ge \frac {2} {3} (a^2+b^2+c^2)</math> | ||
− | After some simplifying, it reduces to <math>a^2+b^2+c^2 \ge ab+bc+ca</math> which is trivial by the Rearrangement Inequality. | + | After some simplifying, it reduces to <math>a^2+b^2+c^2 \ge ab+bc+ca</math> which is trivial by the Rearrangement Inequality. |
==Solution 3== | ==Solution 3== |
Revision as of 15:16, 11 December 2012
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 |