Show that $f(x)=\frac{x}{1+|x|}$ is uniformly continuous. I tried to use the definition and arrived this far:
$|f(x)-f(y)|=\left|\frac{x}{1+|x|}-\frac{y}{1+|y|}\right|=\frac{|x-y+x|y|-y|x||}{(1+|x|)(1+|y|)}\leq|x-y+x|y|-y|x||$.
Any suggestion for ending the proof?
I also tried to prove that $\frac{x}{1+x}$ is uniformly continuous on $[0,\infty[$ and that $\frac{x}{1-x}$ is uniformly continuous on $[-\infty,0[$, but I wonder if we can use the definition with the function $f(x)=\frac{x}{1+|x|}$ itself.

Here is another approach: Your function $f$ is  differeentiable on all of ${\mathbb R}$, whereby $$f'(x)={1\over\bigl(1+|x|\bigr)^2}\qquad(-\infty<x<\infty)\ .$$
As $|f'(x)|\leq1$ for all $x$ the MVT implies that $|f(x)-f(y)|\leq|x-y|$; hence $f$ is even Lipschitz continuous on ${\mathbb R}$.
