Prove $f(x) = \frac{1}{1+x^2}$ is uniformly continuous on $\mathbb{R}$ Prove $f(x) =\frac{1}{1+x^2}$ is uniformly continuous on $\mathbb{R}$.
My attempt...   
Proof 
$$\left| f(x) - f(y) \right| = \left| \frac{1}{1+x^2} - \frac{1}{1+y^2}\right| = \frac{\left|x+y\right|}{\left(1+x^2\right)\left(1+y^2\right)}\left|x-y\right| $$  
By the triangle inequality  
$$\frac{\left|x+y\right|}{\left(1+x^2\right)\left(1+y^2\right)}\left|x-y\right| \leq \left(\frac{\left|x\right|}{\left(1+x^2\right)\left(1+y^2\right)} +\frac{\left|y\right|}{\left(1+x^2\right)\left(1+y^2\right)} \right) \left|x-y\right| \tag{$\star$}$$   
Note that for all $x\in \mathbb{R}$
$$\left|x\right| < 1 +x^2 \implies \left|x\right| < \left(1 +x^2\right)\left(1+y^2\right)$$ 
Therefore 
$$\frac{\left|x\right|}{\left(1 +x^2\right)\left(1+y^2\right)} \leq 1$$  
Applying this fact to $(\star)$ we see that 
$$\left(\frac{\left|x\right|}{\left(1+x^2\right)\left(1+y^2\right)} +\frac{\left|y\right|}{\left(1+x^2\right)\left(1+y^2\right)} \right) \left|x-y\right| \leq \left(1 + 1\right)\left|x-y\right| \leq 2\left|x - y \right|$$
Therefore $f$ is a Lipschitz function, which implies $f$ is uniformly continuous on $\mathbb{R}$. 

Please comment on validity and/or readability, thank you.

Apply mean value theorem in an interval $[x,y], x,y\in \mathbb R$. So, $\exists \xi \in [x,y]: f(x)-f(y)=f'(\xi)(x-y)$. Also we know that $|f'(\xi)|<1$.
