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| − | '''
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| − | 2nd solution:'''
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| − | by RJchan18
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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}</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>.
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| − | Performing the same operation on the two other terms on the left and adding the results together completes the proof.
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| | == Resources == | | == Resources == |
Revision as of 15:42, 14 April 2009
Problem
Let 
, 
, 
be positive real numbers. Prove that
Solution
solution by paladin8:
WLOG, assume 
.
Then the LHS becomes 
.
Notice 
, so 
.
So 
as desired.
Resources
, 
, 
be positive real numbers. Prove that
.
.
, so 
.
as desired.
