how to show that $\frac{s}{1+s^2}$ is Lipschitz continuous? what I've tried so far : 
$$\begin{align}
& s , t \in \mathbb{R} \\
& | F(s) - F(t) |= \frac{|s(1+t^2) - t(1+s^2)|}{|(1+s^2)(1+t^2)|} \leq| s(1+t^2) - t(1+s^2)| \\
& \leq |s-t| + |st||s-t| \\
\end{align}$$
since $s , t \in \mathbb{R} $ I don't know how to proceed anymore, any help will be greatly appreciated.
 A: We have
\begin{align*}
\lvert F(s) - F(t) \rvert
= \frac{\lvert s(1+t^2) - t(1+s^2) \rvert}{(1+s^2)(1+t^2)}
= \frac{\lvert (1-st)(s-t) \rvert}{(1+s^2)(1+t^2)}.
\end{align*}
Now using the AM-GM inequality,
$$ \lvert 1-st \rvert \leq 1+|st| \leq 1+\frac{s^2+t^2}{2} \leq 1+s^2+t^2+s^2t^2 = (1+s^2)(1+t^2), $$
and so, it follows that
$$ \lvert F(s) - F(t) \rvert \leq \lvert s - t \rvert. $$

Addendum. If the mean-value theorem is available to you, the proof can be made a a bit easier (or more routine, to be precise): there exists $\xi$ between $s$ and $t$ such that
$$ \left| F(s) - F(t) \right| = \left| F'(\xi) \right| \left| s - t \right| = \left|\frac{1-\xi^2}{(1+\xi^2)^2} \right| \left|s - t\right|, $$
and it is clear that the prefactor is bounded from above by $1$, hence we get $\leq \left| s - t\right|$.
A: We have:  $|f(s)-f(t)|=\left|\dfrac{s(1+t^2) - t(1+s^2)}{(1+s^2)(1+t^2)}\right|=\left|\dfrac{(s-t)(1-st)}{(1+s^2)(1+t^2)}\right|\le|s-t|\dfrac{\sqrt{(1+s^2)(1+t^2)}}{(1+s^2)(1+t^2)} = \dfrac{|s-t|}{\sqrt{(1+s^2)(1+t^2)}}\le |s-t|$. Here we used Cauchy-Schwarz inequality for $|1-st| = |1\cdot 1+ (-s)t| \le \sqrt{(1^2+(-s)^2)(1+t^2)} = \sqrt{(1+s^2)(1+t^2)}$.
A: Differentiate the function $$ F(s)=\frac {s}{1+s^2}$$ to get $$F'(s)=\frac {1-s^2}{(1+s^2)^2}\le  1$$
$$|F(t)-F(s)|=|F
'(c)(t-s)|\le |t-s|$$
That shows $F$ is Lipschits. 
