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.
 A: What you have ( $2abc+ab+bc+ca=1$ and $a,b,c>0$ ) does actually imply $abc\leq 1/8$. This is because, by AM-GM, $$ab+bc+ca\geq3\sqrt[3]{a^2b^2c^2}$$and so, writing $x=abc$, we have $1\geq 2x+3x^{2/3}.$ 
The RHS is clearly increasing in $x$, and we have equality when $x=1/8$, so any $x>1/8$ violates this inequality.
A: 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!
A: The AM-GM inequality on $2abc$, $ab$, $ac$, $bc$ gives:
$$1 = 2abc+ab+bc+ca \geq 4 \sqrt[4]{2a^3b^3c^3}$$
So $1 \geq 2^{9/4} (abc)^{3/4}$. This gives $2^{-9/4} \geq  (abc)^{3/4}$, so $2^{-3} = 2^{-9/4 \cdot 4/3} \geq  abc$. So $abc \leq \frac18$. 
A: 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.
A: 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.
