If $ { a\over a+1} + { b\over b+1 } + { c\over c+1 } = 1 $ prove $ abc \le 1/8 $ If $ a,b,c $ are positive real numbers and $ { a\over a+1} + {b\over b+1} + {c\over c+1} = 1 $ then prove that $ abc \le {1\over 8} $ 
Could I get some help with this?
What I have tried-
Making the denominators equal and adding to get another equation of the form $ {N\over M}= 1 $ then multiplying by $ M $ both the sides and simplifying to get $ 2abc + ab + bc + ac = 1 $ , from which we have,
 $ abc \lt 1/2 $ 
But which obviously isn't enough to prove that $ abc \le 1/8 $ 
I also tried using Jensen's inequality which states that 
$ f(E(X)) \ge E(f(X)) $ when $ f $ is a concave function and $ f(E(X)) \le E(f(X)) $ when $ f $ is a convex function, and where $ E(X) $ denotes the expectation of X ,
but that didn't really help.

Substitute
\begin{eqnarray*}
a =\frac{x}{y+z} \\
b=\frac{y}{z+x} \\
c= \frac{z}{x+y}
\end{eqnarray*}
This satisfies the constraint. Now the following quantity is clearly positive
\begin{eqnarray*}
x(y-z)^2+y(z-x)^2+z(x-y)^2 \geq 0 \\
(x+y)(y+z)(z+x) \geq 8xyz \\
\end{eqnarray*}
Rearrange a bit & substitute back in & the result follows.
