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.

If you multiply both sides by $(a+1)(b+1)(c+1)$ you'll obtain $$3abc+2ab+2ac+2bc+a+b+c=(a+1)(b+1)(c+1),$$
which simplifies to $$3abc+2ab+2ac+2bc+a+b+c=1 + a + b + a b + c + a c + b c + a b c,$$
rearranging to $$abc=\frac{1-(ab+ac+bc)}{2}.$$
Using AM-GM you get
$$1-2abc=ab+ac+bc\geq3\sqrt[3]{a^2b^2c^2}.$$ Now set $u=abc$ and you have $$u=\frac{1-\sqrt[3]{u^2}}{2}$$ which leads to $$1\geq2u+3u^{2/3}.$$
Now you see that $u=abc>1/8$ violates the inequality.
