RMO 2016

then why did you post a solution?

This is the best I can do... If you're looking for a real solution to this problem, then skip over this comment.

Using the method of Lagrange multipliers, I get the equations:

$bc=\lambda(\frac{1}{1+b}-\frac{c}{(1+a)^2})$

$ac=\lambda(\frac{1}{1+c}-\frac{a}{(1+b)^2})$

$ab=\lambda(\frac{1}{1+a}-\frac{b}{(1+c)^2}).$

I next note that $a=b=c$ solves the equations, giving consistent values for $\lambda.$

Substituting this into $\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+a}=1$ yields $a=b=c=\frac{1}{2}.$

So, I find $\frac{1}{8}$ is a possible value for the maximum of $abc$ .

The point here is to show that $\frac{1}{8}$ is the global maximum.

If someone figures out how to show this, please comment here. Thanks!