Difference between revisions of "2003 USAMO Problems/Problem 5"
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== Solution == | == Solution == | ||
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− | + | Since all terms are homogeneous, we may assume WLOG that <math>a + b + c = 3</math>. | |
Then the LHS becomes <math>\sum \frac {(a + 3)^2}{2a^2 + (3 - a)^2} = \sum \frac {a^2 + 6a + 9}{3a^2 - 6a + 9} = \sum \left(\frac {1}{3} + \frac {8a + 6}{3a^2 - 6a + 9}\right)</math>. | Then the LHS becomes <math>\sum \frac {(a + 3)^2}{2a^2 + (3 - a)^2} = \sum \frac {a^2 + 6a + 9}{3a^2 - 6a + 9} = \sum \left(\frac {1}{3} + \frac {8a + 6}{3a^2 - 6a + 9}\right)</math>. | ||
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Notice <math>3a^2 - 6a + 9 = 3(a - 1)^2 + 6 \ge 6</math>, so <math>\frac {8a + 6}{3a^2 - 6a + 9} \le \frac {8a + 6}{6}</math>. | Notice <math>3a^2 - 6a + 9 = 3(a - 1)^2 + 6 \ge 6</math>, so <math>\frac {8a + 6}{3a^2 - 6a + 9} \le \frac {8a + 6}{6}</math>. | ||
− | So <math>\sum \frac {(a + 3)^2}{2a^2 + (3 - a)^2} \le \sum \left(\frac {1}{3} + \frac {8a + 6}{6}\right) = 1 + \frac {8(a + b + c) + 18}{6} = 8</math> as desired. | + | So <math>\sum \frac {(a + 3)^2}{2a^2 + (3 - a)^2} \le \sum \left(\frac {1}{3} + \frac {8a + 6}{6}\right) = 1 + \frac {8(a + b + c) + 18}{6} = 8</math>, as desired. |
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== Resources == | == Resources == |
Revision as of 20:32, 13 April 2012
Problem
Let ,
,
be positive real numbers. Prove that
![$\dfrac{(2a + b + c)^2}{2a^2 + (b + c)^2} + \dfrac{(2b + c + a)^2}{2b^2 + (c + a)^2} + \dfrac{(2c + a + b)^2}{2c^2 + (a + b)^2} \le 8.$](http://latex.artofproblemsolving.com/d/2/b/d2be8552ac3b2dcfb8d235a80ddc4d812b2f2155.png)
Solution
Since all terms are homogeneous, we may assume WLOG that .
Then the LHS becomes .
Notice , so
.
So , as desired.