Lipschitz continuity of a function Show that $g(x) = \frac{1}{x^{2} +1}$ is Lipschitz conitnuous.

From the definition, we must show that $\forall x,y \in \mathbb{R}$, $|f(x)-f(y)| \leq K|x-y|$, for some real constant $K$.
First, I tried to find:
$|f(x)-f(y)|=|\displaystyle\frac{1}{x^{2} +1} - \frac{1}{y^{2} +1}|=|\frac{x^2-y^2}{(1+x^2)(1+y^2)}| \leq \frac{(|x-y|)(|x|+|y|)}{(1+x^2)(1+y^2)}$
Using the triangle inequality. However, I hit a dead end here. What would be the best way to approach this problem? Thank you for the help.

HINT:
$$
\frac{(|x|+|y|)}{(1+x^2)(1+y^2)}\leq2.
$$
For $|x|$, $|y|\leq1$ it is obvious. If $|x|>1$, $x^2>|x|$, and $\frac{|x|}{1+x^2}\leq1$; similarly for $y$
