$\frac{x}{1+x}=1-\frac{1}{1+x}$ so for any $x,y,z$ if we look at $f(x)+f(y)>f(z)$ the exporession is equivalent to $1+\frac{1}{1+z}$> $\frac{1}{1+x}$+$\frac{1}{1+y}$. Cleaning things up we have to prove that : $2xy+xyz+x+y>z$ which is clearly true.

See here and here.