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Got a problem proving inequality Rate Topic: -----

#1 4 December 2011 - 08:00 PM Ren 


Lepton
Let a,b, and c be positive real numbers. Prove that
\sqrt{\frac {a}{a+b}}+ \sqrt{\frac {b}{b+c}}+ \sqrt{\frac {c}{c+a}} \leq 2\sqrt{1+\frac {abc}{(a+b)(b+c)(c+a)}}

I've tried squaring both sides, and that eventually lead me to nothing. Then I tried taking the common denominator of the left side, but I got nothing either... There has to be some trick involved that I'm not seeing... Anyone has any suggestions or input, that will be great!
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#2 4 December 2011 - 10:26 PM the tree 


Primate
Try making these substitutions: \left( \frac{a}{a+b} \right) =x^2 , \, \left( \frac{b}{b+c} \right)=y^2 , \, \left( \frac{c}{c+a}\right)=z^2

It should be fairly easy from there.

This post has been edited by the tree: 4 December 2011 - 10:27 PM

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#3 4 December 2011 - 11:45 PM uncool 


Atom
the tree: I don't think it's quite so simple just from there. There is a constraint on x^2, y^2, and z^2 that is necessary to prove the relationship.

Namely, ((1/x^2) - 1)((1/y^2) - 1)((1/z^2) - 1) = 1.
=Uncool-
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