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Difference between revisions of "2003 USAMO Problems/Problem 5"
(another solution to this problem)
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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.
 
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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2nd solution:
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Because this inequality is symmetric, let's examine the first term on the left side of the inquality.
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let <math>x=a+b, y=a+c</math> and <math>z=b+c</math>. So <math>\frac{(2a+b+c)^2}{2a^2+(b+c)^2}=\frac{(x+y)^2}{(x+y-z)^2+z^2}</math>.
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Note that <math>(x+y-z)+(z)=x+y</math>. So Let <math>(x+y-z)=m</math>, <math>x+y=m+z</math>. QM-AM gives us <math>\sqrt\frac{m^2+z^2}{2)</math> <math>\geq \frac{m+z}{2}. Squaring both sides and rearranging the inequality gives us </math>\frac{m^2+z^2}{(m+z)^2}\geq \frac{1}{2}<math> so </math>\frac{(m+z)^2}{m^2+z^2}\leq 2<math> so </math>\frac{(x+y)^2}{(x+y-z)^2+z^2}\leq 2<math> thus </math>\frac{(2a+b+c)^2}{2a^2+(b+c)^2}\leq 2$.
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Performing the same operation on the two other terms on the left and adding the results together completes the proof.
  
 
== Resources ==
 
== Resources ==

Revision as of 14:26, 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:

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.

Resources

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