Let a, b, and c be the sides of a triangle. Prove that $(a+b-c)(b+c-a)(c+a-b)\leq abc$ 
Possible Duplicate:
Prove $(-a+b+c)(a-b+c)(a+b-c) \leq abc$, where $a, b$ and $c$ are positive real numbers 

The title says it all. I've been trying to prove this for hours! Help me!

What I did was wrong.
Use  the substitution $a=x+y$, $b=y+z$ and $c=z+x$ in the initial inequality to have $$8xyz\le(x+y)(y+z)(z+x)$$ and that follows directly from $AM-GM$ since 
$$\begin{array}{ll}x+y\ge 2\sqrt{xy}\\ y+z\ge 2\sqrt{yz}\\ z+x\ge 2\sqrt{zx}\\(x+y)(y+z)(z+x)\ge 8\sqrt{x^2y^2z^2}=8xyz\\  \end{array}$$
NB: $AM-GM$ applies only because the sides of a triangle are positive, thus $a,b,c>0$.
