$z ≤ x + y$ 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?
 A: From what you have left off: $z + zx + zy + xyz \le x+xz + y+ yz + 2xy + 2xyz\iff z \le x+y + 2xy+xyz $, and this is true since $z \le x+y$, and also $0 \le 2xy$, $0 \le xyz$. Thus your proof is true, but you need to expand it to show all the little details...
A: If
$z \le \sum_{i=1}^n x_i
$
then
we want to show that
$\dfrac{z}{1+z}
\le \sum_{i=1}^n \dfrac{x_i}{1+x_i}
$.
Let
$f(z)
=\dfrac{z}{1+z}
$.
We want to show that
$f(z)
\le \sum_{i=1}^n f(x_i)
$.
$f(z)
=\dfrac{z}{1+z}
=1-\dfrac{1}{1+z}
$
so
$f'(z)
=\dfrac1{(1+z)^2} 
> 0
$.
$f$ is increasing so
$f(z)
\le f(\sum_{i=1}^n x_i)
$.
Therefore,
if we can show that
$f(\sum_{i=1}^n x_i)
\le \sum_{i=1}^n f(x_i)
$
we are done.
For $n=2$,
the difference is
$\begin{array}\\
f(x)+f(y)-f(x+y)
&=\dfrac{x}{1+x}+\dfrac{y}{1+y}-\dfrac{x+y}{1+x+y}\\
&=\dfrac{x}{1+x}+\dfrac{y}{1+y}-\dfrac{x}{1+x+y}-\dfrac{y}{1+x+y}\\
&=\dfrac{x}{1+x}-\dfrac{x}{1+x+y}+\dfrac{y}{1+y}-\dfrac{y}{1+x+y}\\
&=x(\dfrac{1}{1+x}-\dfrac{1}{1+x+y})+y(\dfrac{1}{1+y}-\dfrac{1}{1+x+y})\\
&=x\dfrac{(1+x+y)-(1+x)}{(1+x)(1+x+y)})+y\dfrac{(1+x+y)-(1+y)}{(1+y)(1+x+y)})\\
&=x\dfrac{y}{(1+x)(1+x+y)})+y\dfrac{x}{(1+y)(1+x+y)})\\
&=xy(\dfrac{1}{(1+x)(1+x+y)}+\dfrac{1}{(1+y)(1+x+y)})\\
&\ge 0\\
\end{array}
$
This easily generalizes.
$\begin{array}\\
\sum_{i=1}^n f(x_i)-f(\sum_{i=1}^n x_i)
&=\sum_{i=1}^n \dfrac{x_i}{1+x_i}-\dfrac{\sum_{i=1}^n x_i}{1+\sum_{i=1}^n x_i}\\
&=\sum_{i=1}^n \dfrac{x_i}{1+x_i}-\sum_{i=1}^n\dfrac{ x_i}{1+\sum_{j=1}^n x_j}\\
&=\sum_{i=1}^n (\dfrac{x_i}{1+x_i}-\dfrac{ x_i}{1+\sum_{j=1}^n x_j})\\
&=\sum_{i=1}^n x_i(\dfrac{1}{1+x_i}-\dfrac{ 1}{1+\sum_{j=1}^n x_j})\\
&=\sum_{i=1}^n x_i(\dfrac{(1+\sum_{j=1}^n x_j)-(1+x_i)}{(1+x_i)(1+\sum_{j=1}^n x_j)})\\
&=\sum_{i=1}^n x_i(\dfrac{\sum_{j=1,j\ne i}^n x_j}{(1+x_i)(1+\sum_{j=1}^n x_j)})\\
&\ge 0
\qquad\text{since each term is }\ge 0\\
\end{array}
$
and we are done.
Since
$f''(z)
=-\dfrac{2}{(1+z)^3}
< 0$,
by Jensen's inequality
we have
$f(\frac1{n}\sum x_i)
\ge \frac1{n}\sum f(x_i)
$
so that
$\sum f(x_i)
\le nf(\frac1{n}\sum x_i)
$.
A: 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 $$\dfrac{z}{1+z} <= \dfrac{x+y}{1+x+y} <= \dfrac{x}{1+x} + \dfrac{y}{1+y}$$  
