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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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2nd solution:'''
  
 
Because this inequality is symmetric, let's examine the first term on the left side of the inquality.  
 
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>.
 
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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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}</math>.  
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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</math>.
  
 
Performing the same operation on the two other terms on the left and adding the results together completes the proof.
 
Performing the same operation on the two other terms on the left and adding the results together completes the proof.

Revision as of 14:27, 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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