Difference between revisions of "2003 USAMO Problems/Problem 5"
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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>. | ||
− | 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 | + | 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>. |
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>. | 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>. |
Revision as of 15:32, 14 April 2009
Problem
Let ,
,
be positive real numbers. Prove that
![$\dfrac{(2a + b + c)^2}{2a^2 + (b + c)^2} + \dfrac{(2b + c + a)^2}{2b^2 + (c + a)^2} + \dfrac{(2c + a + b)^2}{2c^2 + (a + b)^2} \le 8.$](http://latex.artofproblemsolving.com/d/2/b/d2be8552ac3b2dcfb8d235a80ddc4d812b2f2155.png)
Solution
solution by paladin8:
WLOG, assume .
Then the LHS becomes .
Notice , so
.
So as desired.
2nd solution:
Because this inequality is symmetric, let's examine the first term on the left side of the inquality.
let and
. So
.
Note that . So Let
,
. QM-AM gives us $\sqrt{\frac{m^2+z^2}{2}$ (Error compiling LaTeX. Unknown error_msg)
.
Squaring both sides and rearranging the inequality gives us so
so
thus
.
Performing the same operation on the two other terms on the left and adding the results together completes the proof.