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Difference between revisions of "2003 USAMO Problems/Problem 5"

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== Solution ==
 
== Solution ==
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{{solution}}
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solution by paladin8:
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WLOG, assume <math>a + b + c = 3</math>.
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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>.
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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.
  
 
== Resources ==
 
== Resources ==
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* [[2003 USAMO Problems]]
 
* [[2003 USAMO Problems]]
 
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53672 Discussion on AOPS/MathLinks]
 
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53672 Discussion on AOPS/MathLinks]
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* [http://www.mathlinks.ro/Forum/viewtopic.php?t=46 Discussion 2]
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* [http://www.mathlinks.ro/Forum/viewtopic.php?t=28678 Discussion 3]
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* [http://www.mathlinks.ro/Forum/viewtopic.php?t=43278 Discussion 4]
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* [http://www.mathlinks.ro/Forum/viewtopic.php?t=48989 Discussion 5]
  
 
[[Category:Olympiad Inequality Problems]]
 
[[Category:Olympiad Inequality Problems]]

Revision as of 18:58, 18 December 2008

Problem

Let $a$, $b$, $c$ be positive real numbers. Prove that

$\dfrac{(2a + b + c)^2}{2a^2 + (b + c)^2} + \dfrac{(2b + c + a)^2}{2b^2 + (c + a)^2} + \dfrac{(2c + a + b)^2}{2c^2 + (a + b)^2} \le 8.$

Solution

solution by paladin8:

WLOG, assume $a + b + c = 3$.

Then the LHS becomes $\sum \frac {(a + 3)^2}{2a^2 + (3 - a)^2} = \sum \frac {a^2 + 6a + 9}{3a^2 - 6a + 9} = \sum \left(\frac {1}{3} + \frac {8a + 6}{3a^2 - 6a + 9}\right)$.

Notice $3a^2 - 6a + 9 = 3(a - 1)^2 + 6 \ge 6$, so $\frac {8a + 6}{3a^2 - 6a + 9} \le \frac {8a + 6}{6}$.

So $\sum \frac {(a + 3)^2}{2a^2 + (3 - a)^2} \le \sum \left(\frac {1}{3} + \frac {8a + 6}{6}\right) = 1 + \frac {8(a + b + c) + 18}{6} = 8$ as desired.

Resources

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