<s>[INST] 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.
 [/INST] \begin{align}
\left|\frac{x^2-y^2}{(1+x^2)(1+y^2)}\right|&=\left|\frac{x+y}{(1+x^2)(1+y^2)}\right| |x-y|\\
&\le \max\left(\frac{x+y}{(1+x^2)(1+y^2)}\right)|x-y|\\
&={3\sqrt{3}\over 8}\,|x-y|
\end{align}
where the last equality follows from a little calculus.
Of course, we do not need such a sharp $K$, so we can of course estimate $$\left|\frac{x+y}{(1+x^2)(1+y^2)}\right|$$ instead as outlined in other solutions.
</s>