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

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== Solution ==
 
== Solution ==
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solution by paladin8:
 
  
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WLOG, assume <math>a + b + c = 3</math> since all terms are homogeneous.
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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>.
  
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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.
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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.
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== Resources ==
 
== Resources ==

Revision as of 20:32, 13 April 2012

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

Since all terms are homogeneous, we may assume WLOG that $a + b + c = 3$.

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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