Show $1/(1+ x^2)$ is uniformly continuous on $\Bbb R$. 
Prove that the function $x \mapsto \dfrac 1{1+ x^2}$ is uniformly continuous on $\mathbb{R}$.

Attempt: By definition a function $f: E →\Bbb R$ is uniformly continuous iff for every $ε > 0$, there is a $δ > 0$ such that $|x-a| < δ$ and $x,a$ are elements of $E$ implies $|f(x) - f(a)| < ε.$
Then suppose $x, a$ are elements of $\Bbb R. $
Now
\begin{align}
|f(x) - f(a)| 
&= \left|\frac1{1 + x^2} - \frac1{1 + a^2}\right|
\\&= \left|  \frac{a^2 - x^2}{(1 + x^2)(1 + a^2)}\right|
\\&= |x - a| \frac{|x + a|}{(1 + x^2)(1 + a^2)}
\\&≤ |x - a| \frac{|x| + |a|}{(1 + x^2)(1 + a^2)}
\\&= |x - a| \left[\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)}\right]
\end{align}
I don't know how to simplify more. Can someone please help me finish? Thank very much.

$$x^2 \geq 0 \implies 1+x^2 > 1 \implies \frac{1}{1+x^2} < 1$$
Using the above inequality,
$$
\begin{align}
|f(x)-f(a)| &\leq |x - a| \frac{|x + a|}{(1 + x^2)(1 + a^2)}\\
&\leq |x - a||x+a|\\
&\leq |x - a|(|x|+|a|)
\end{align}
$$
Choose $$|x-a| \leq 1 \implies |x|\leq |a|+1$$
Now,
$$
\begin{align}
|f(x)-f(a)| &\leq |x - a|(|x|+|a|)\\
&\leq |x-a|(2|a|+1)
\end{align}
$$
Choose $$|x-a| < \frac{\epsilon}{2|a|+1} $$
Now,
$$
\begin{align}
|f(x)-f(a)| &\leq \epsilon
\end{align}
$$
where $$\delta = Inf\{1, \frac{\epsilon}{2|a|+1}\}$$
