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

Revision as of 14:32, 14 April 2009

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.

2nd solution:

by RJchan18

Because this inequality is symmetric, let's examine the first term on the left side of the inquality.

let $x=a+b, y=a+c$ and $z=b+c$ . So $\frac{(2a+b+c)^2}{2a^2+(b+c)^2}=\frac{(x+y)^2}{(x+y-z)^2+z^2}$ .

Note that $(x+y-z)+(z)=x+y$ . So Let $(x+y-z)=m$ , $x+y=m+z$ . QM-AM gives us $\sqrt{\frac{m^2+z^2}{2}$ (Error compiling LaTeX. Unknown error_msg) $\geq \frac{m+z}{2}$ .

Squaring both sides and rearranging the inequality gives us $\frac{m^2+z^2}{(m+z)^2}\geq \frac{1}{2}$ so $\frac{(m+z)^2}{m^2+z^2}\leq 2$ so $\frac{(x+y)^2}{(x+y-z)^2+z^2}\leq 2$ thus $\frac{(2a+b+c)^2}{2a^2+(b+c)^2}\leq 2$ .

Performing the same operation on the two other terms on the left and adding the results together completes the proof.

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

Revision as of 14:32, 14 April 2009

Problem

Solution

Resources

@@ Line 19: / Line 19: @@
 '''
 nd solution:'''
+by RJchan18
 Because this inequality is symmetric, let's examine the first term on the left side of the inquality.