Difference between revisions of "2012 USAJMO Problems/Problem 3"
(→Solution 2=) |
(→Solution 2) |
||
| Line 29: | Line 29: | ||
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> | ||
| − | Then use Titu's Lemma | + | 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. | ||
Revision as of 16:53, 25 April 2012
Contents
Problem
Let 
, 
, 
be positive real numbers. Prove that
![\[\frac{a^3 + 3b^3}{5a + b} + \frac{b^3 + 3c^3}{5b + c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{2}{3} (a^2 + b^2 + c^2).\]](http://latex.artofproblemsolving.com/4/4/8/448b311f5688c5c18e520bfe1a9c56906700c3d2.png)
Solution
By the Cauchy-Schwarz inequality,
![\[[a(5a + b) + b(5b + c) + c(5c + a)] \left( \frac{a^3}{5a + b} + \frac{b^3}{5b + c} + \frac{c^3}{5c + a} \right) \ge (a^2 + b^2 + c^2)^2,\]](http://latex.artofproblemsolving.com/c/8/5/c853ed04ce97a5d53eb2c179cc962c2d9e0af755.png)
so
![\[\frac{a^3}{5a + b} + \frac{b^3}{5b + c} + \frac{c^3}{5c + a} \ge \frac{(a^2 + b^2 + c^2)^2}{5a^2 + 5b^2 + 5c^2 + ab + ac + bc}.\]](http://latex.artofproblemsolving.com/c/7/0/c70b85f9d08ce18cabc92505778450fa22084b5d.png)
Since 
,
![\[\frac{(a^2 + b^2 + c^2)^2}{5a^2 + 5b^2 + 5c^2 + ab + ac + bc} \ge \frac{(a^2 + b^2 + c^2)^2}{6a^2 + 6b^2 + 6c^2} = \frac{1}{6} (a^2 + b^2 + c^2).\]](http://latex.artofproblemsolving.com/c/6/7/c67539d05946214142c90df1baebf0bcc6fc3875.png)
Hence,
![\[\frac{a^3}{5a + b} + \frac{b^3}{5b + c} + \frac{c^3}{5c + a} \ge \frac{1}{6} (a^2 + b^2 + c^2).\]](http://latex.artofproblemsolving.com/2/5/b/25b8b46bbae4a04cad9f782369cbd6b469179254.png)
Again by the Cauchy-Schwarz inequality,
![\[[b(5a + b) + c(5b + c) + a(5c + a)] \left( \frac{b^3}{5a + b} + \frac{c^3}{5b + c} + \frac{a^3}{5c + a} \right) \ge (a^2 + b^2 + c^2)^2,\]](http://latex.artofproblemsolving.com/d/f/4/df4ba3115174961740e725bbf3e055ee3c95ca60.png)
so
![\[\frac{b^3}{5a + b} + \frac{c^3}{5b + c} + \frac{a^3}{5c + a} \ge \frac{(a^2 + b^2 + c^2)^2}{a^2 + b^2 + c^2 + 5ab + 5ac + 5bc}.\]](http://latex.artofproblemsolving.com/a/0/9/a091852266f2a5080dcdc2d9bd4f5aabb253c303.png)
Since ,
![\[\frac{(a^2 + b^2 + c^2)^2}{a^2 + b^2 + c^2 + 5ab + 5ac + 5bc} \ge \frac{(a^2 + b^2 + c^2)^2}{6a^2 + 6b^2 + 6c^2} = \frac{1}{6} (a^2 + b^2 + c^2).\]](http://latex.artofproblemsolving.com/0/c/5/0c593c978f6e27800e4421768d0df18ee5d7ee1d.png)
Hence,
![\[\frac{b^3}{5a + b} + \frac{c^3}{5b + c} + \frac{a^3}{5c + a} \ge \frac{1}{6} (a^2 + b^2 + c^2).\]](http://latex.artofproblemsolving.com/5/4/a/54abb3c7c80aa1733717f8d78fa9680a3cf52dc2.png)
Therefore,
![\[\frac{a^3 + 3b^3}{5a + b} + \frac{b^3 + 3c^3}{5b + c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{1 + 3}{6} (a^2 + b^2 + c^2) = \frac{2}{3} (a^2 + b^2 + c^2).\]](http://latex.artofproblemsolving.com/9/7/3/9736e742df8602ada135197c276e1cdd5c4d108c.png)
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.
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 | ||
