Got a problem

[Ren]

Let a,b, and c be positive real numbers. Prove that

[math]\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)}}[/math]

 

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!

[the tree]

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

 

It should be fairly easy from there.

Edited December 4, 2011 by the tree

[uncool]

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-