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 $x+c$.  
 A: Try this: let $x,y > 0$ (for simplicity) be two positive reals. Then set
$$
f(x) = \frac{1}{1 + x^2}
$$
so that
$$
f(x) - f(y) = \frac{y^2 - x^2}{(1 + x^2)(1 + y^2)} = (y - x){y + x \over (1 + x^2) (1 + y^2)}
$$
Thus
$$
|f(x) - f(y)| \leq |x - y| \cdot \left| {y + x \over (1 + x^2)(1 + y^2)}\right|
$$
The right-hand factor can be bounded from above by
$$
\left| \frac{x}{1 + x^2}\right| + \left| \frac{y}{1 + y^2}\right|
$$
Can these terms get big as $x,y$ range over $\mathbb{R}$? How do you conclude uniform continuity for $f$?
A: Here is one way.
Let $f(x) = \frac{1}{1+x^2}$. Then $f'(x) = - \frac{2x}{(1+x^2)^2}$. A simple estimate (split into $|x|<1$ and $|x| \ge 1$) shows $|f'(x)|\le 2$ for all $x$.
Then the mean value theorem gives (for some $\xi$ depending on $x,y$): $|f(x)-f(y)| = |f'(\xi)| |x-y| \le 2 |x-y|$, from we see that $f$ is Lipschitz continuous and hence uniformly continuous.
Here is another way:
Let $\epsilon>0$. Choose $L$ such that if $|x|>L$, then $|f(x)| < \frac{1}{2} \epsilon$. Then if $|x|>L$, $|y|>L$, we have $|f(x)-f(y)| < \frac{1}{2} \epsilon + \frac{1}{2} \epsilon = \epsilon$.
Since $[-2L,2L]$ is compact, the continuous function $f$ is uniformly continuous on $[-2L,2L]$, and so there exists some $\delta>0$ such that if $|x-y| < \delta$ (and $x,y \in [-2L,2L]$), then $|f(x)-f(y)| < \epsilon$.
It follows that if $|x-y|< \delta$ (with $x,y \in \mathbb{R}$), that $|f(x)-f(y)| < \epsilon$, and so
$f$ is uniformly continuous on $\mathbb{R}$.
A: 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 $\delta = \epsilon /\alpha$.
