Q:

Show that $|h(x)-h(y)|≤|x-y|$

Let $$h(x)=x/(1+|x|)$$ 
Show that $$|h(x)-h(y)|≤|x-y|$$ 
I have no idea to start. But I can see that $h(x)≤x$. Logically, the required inequality will follow from this. But I cannot apply it to $y$.

A:

Suppose that $0 < x < y$. You have $$\vert h(x)-h(y) \vert = \left \vert\frac{x}{1+x}-\frac{y}{1+y}\right\vert=\frac{\vert x-y \vert}{(1+x)(1+y)} \le \vert x -y \vert.$$
Then study the case $x < y <0$. And finally the last one $x < 0 < y$

