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

Revision as of 09:57, 10 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

solution by paladin8:

WLOG, assume $a + b + c = 3$ since all terms are homogeneous.

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

@@ Line 6: / Line 6: @@
 solution by paladin8:
-WLOG, assume <math>a + b + c = 3</math>.
+WLOG, assume <math>a + b + c = 3</math> since all terms are homogeneous.
 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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Difference between revisions of "2003 USAMO Problems/Problem 5"

Revision as of 09:57, 10 April 2012

Problem

Solution

Resources