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.
 A: Hint: $$\frac{|x|}{(1+x^2)(1+a^2)} \leq  \frac{|x|}{1+x^2} < 1$$
and
$$\frac{|a|}{(1+x^2)(1+a^2)} \leq  \frac{|a|}{1+a^2} < 1$$
A: Hint: use the inequality
$$x>0\implies 
x < 1 + x^2
$$
(if $x<1$ it is true; otherwise via multiplication by $x$, $x>1\implies x^2>x$)
A: First:$f(x)=\frac{1}{1+x^2}\implies f'(x)=-\frac{2x}{(1+x^2)^2} $
Then observes that; For $|x|\le1$ 
$$\frac{|x|}{(1+x^2)^2} \le\frac{1}{(1+x^2)^2}\le  1$$
and for $|x|\ge1$ 
$$|x|\le x^2 \le (1+x^2)^2\implies \frac{|x|}{(1+x^2)^2} \le 1$$
Hence,
$$|f'(x)|=\frac{2|x|}{(1+x^2)^2} \le 2\implies |f(x)-f(y)|\le 2|x-y|$$

This shows that $f$ is Lipschitz which implies, that $f$ is uniformly continuous.

A: You are nearly finishing the proof.
$$|x - a| (\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)})\le |x - a| (\frac{1}{2(1 + a^2)} + \frac{1}{2(1 + x^2)})\le |x-a|$$
Take $\delta=\epsilon$.
A: According to the mean value theorem,
$$
\left|\frac1{1 + x^2} - \frac1{1 + a^2}\right| = f'(c)|x-a|
$$
where $f(x)=\dfrac 1 {1+x^2}$ and $c$ is somewhere between $x$ and $a$.  But $|f'(c)|\le\max |f'|$, the absolute maximum value of $|f'|$.  In order for this to make sense, you need to show that $|f'|$ does have an absolute maximum value.  But that is not hard.  So you have
$$
|f(x)-f(a)|\le M|x-a| \text{ for ALL values of $x$ and $a$},
$$
(where $M$ is the absolute maximum of $|f'|$).  So $f$ is Lipschitz-continuous and therefore uniformly continuous.
A: $$x^2 \geq 0 \implies 1+x^2 > 1 \implies \frac{1}{1+x^2} < 1$$
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 $$\delta = Inf\{1, \frac{\epsilon}{2|a|+1}\}$$
