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

Revision as of 12:39, 4 July 2013

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.

@@ Line 19: / Line 19: @@
 [[Category:Olympiad Algebra Problems]]
 [[Category:Olympiad Inequality Problems]]
+{{MAA Notice}}

2003 USAMO (Problems • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

Page

Toolbox

Search

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

Revision as of 12:39, 4 July 2013

Problem

Solution

See also