Simple Uniform Continuity Show that:
$\cfrac1{x^2+1}$ is uniformly continuous on $\mathbb R$!
I'm having difficulty finding an inequality between $1+x^2+c^2+x^2c^2$ and $x+c$.  

Hint: Let $f$ be the relevant function, and $c$ a small number. Then
\begin{align} |f(x + c) - f(x)| &= \left|\frac{1}{(x + c)^2 + 1} - \frac{1}{x^2 + 1}\right| \\
&= \left|\frac{x^2 + 1  - ((x + c)^2 + 1)}{((x + c)^2 + 1)(x^2 + 1)}\right| \\
&= \frac{|2cx + c^2|}{x^2 + 1} \\
&= |c| \frac{|2x + c|}{x^2 + 1} \\
&\le |c| \frac{|2x|}{x^2 + 1} + \frac{|c|^2 }{x^2 + 1} \\
&\le |c| \frac{|2x|}{x^2 + 1} + |c| 
\end{align}
provided $|c| \le 1$. Now show that the quantity $\frac{|2x|}{x^2 + 1}$ is bounded by some fixed constant $\alpha$, and complete the proof using $\delta = \epsilon /\alpha$.
