Show that the function $f(x)=\frac1{1+x^2}$ satisfies a global Lipschitz condition on $\mathbb R$.
By the definition we want to whow that $|f(x)-f(y)|\leqslant L|x-y|$. Here is my work so far. $$\left|\frac1{1+x^2}-\frac1{1+y^2}\right|=\left|\frac{y^2-x^2}{(1+y^2)(1+x^2)}\right|\leqslant L|x-y|$$ $$\implies\left|\frac{y+x}{(1+y^2)(1+x^2)}\right|\leqslant L.$$ It appears to be bounded, since the denominator becomes much larger than the numerator for large values of $x$ and $y$. Is this good enough or can I find a value for $L$? Also, is it also the same as showing that $f'(x)$ is globally bounded?
Thank you in advance for any help or constructive criticism. I am forever grateful.