Globally Lipschitz condition 
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.

$$\left| \frac{x+y}{(1+x^2)(1+y^2)}\right|=\frac{|x+y|}{(1+x^2)(1+y^2)}$$
By triangle inequality:
$$\frac{|x+y|}{(1+x^2)(1+y^2)} \leq \frac{|x|}{(1+x^2)(1+y^2)}+\frac{|y|}{(1+x^2)(1+y^2)}$$
Now $x^2+1 \geq 1$, so:
$$ \frac{|x|}{(1+x^2)(1+y^2)} \leq \frac{|x|}{1+x^2}$$
Next $\displaystyle \frac{|x|}{1+x^2} \leq 2$ be AM-GM inequality. The same with $\displaystyle \frac{|y|}{(1+x^2)(1+y^2)}$, so finally:
$$\left| \frac{x+y}{(1+x^2)(1+y^2)}\right| \leq 4$$
