$z ≤ x + y$ implies $z/(1 + z) ≤ x/(1 + x) + y/(1 + y)$ 
Suppose $x,y,z $ be nonnegative reals. Show that $z ≤ x + y\implies z/(1 + z) ≤ x/(1 + x) + y/(1 + y)$.

My Proof: If $z=0$, then we are done. So, suppose $z>0$. Since $z ≤ x + y$, and $x$ and $y$ nonnegative, $z ≤ x + y+2xy+xyz$, which leads to $z(1+x+y+xy)\le (x+y+2xy)(1+z)$ and since $z\ne 0$, we get $z/(1 + z) ≤x/(1 + x) + y/(1 + y)$.
Is my proof correct?

Yes your proof is correct. I have provided another proof as follows.  Notice that $f(t) = \dfrac{t}{1+t}$ is an increasing function on its domain. Therefore if $x$, $y$, and $z$ are non-negative real numbers and $z$ is less than or equal to $x+y$ then f(z) is less than or equal to f(x+y). That is $$\dfrac{z}{1+z} <= \dfrac{x+y}{1+x+y} <= \dfrac{x}{1+x} + \dfrac{y}{1+y}$$  
