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

(New page: == Problem == Let <math>a</math>, <math>b</math>, <math>c</math> be positive real numbers. Prove that <center><math>\dfrac{(2a + b + c)^2}{2a^2 + (b + c)^2} + \dfrac{(2b + c + a)^2}{2b^2 ...)
 
m (Problem)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 
 
Let <math>a</math>, <math>b</math>, <math>c</math> be positive real numbers. Prove that
 
Let <math>a</math>, <math>b</math>, <math>c</math> be positive real numbers. Prove that
 
<center><math>\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.</math></center>
 
<center><math>\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.</math></center>

Revision as of 10:52, 8 October 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

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