<s>[INST] 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.
 [/INST] By AM-GM:$$\frac{1}{a+1}=1-\frac{a}{a+1}=\frac{b}{b+1}+\frac{c}{c+1}\geq2\sqrt{\frac{bc}{(b+1)(c+1)}}.$$
By the permutation symmetry, the inequality is invariant with respect to the cyclic permutation of $(a,b,c)$.
Thus, $$\frac{1}{(a+1)(b+1)(c+1)}\geq\frac{8abc}{(a+1)(b+1)(c+1)}$$
and we are done!
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