Difference between revisions of "2012 USAJMO Problems/Problem 3"
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<cmath>\frac{a^3 + 3b^3}{5a + b} + \frac{b^3 + 3c^3}{5b + c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{1 + 3}{6} (a^2 + b^2 + c^2) = \frac{2}{3} (a^2 + b^2 + c^2).</cmath> | <cmath>\frac{a^3 + 3b^3}{5a + b} + \frac{b^3 + 3c^3}{5b + c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{1 + 3}{6} (a^2 + b^2 + c^2) = \frac{2}{3} (a^2 + b^2 + c^2).</cmath> | ||
− | =Solution 2== | + | ==Solution 2== |
Split up the numerators and multiply both sides of the fraction by either a or 3a: | Split up the numerators and multiply both sides of the fraction by either a or 3a: | ||
<math>\sum_{cyc} \frac {a^4} {5a^2+ab} +\sum_{cyc} \frac {9a^4} {15ac+3a^2}</math> | <math>\sum_{cyc} \frac {a^4} {5a^2+ab} +\sum_{cyc} \frac {9a^4} {15ac+3a^2}</math> |
Revision as of 16:52, 25 April 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 (like Cauchy): It suffices to prove that After some simplifying, it reduces to which is trivial by the Rearrangement Inequality.
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 |