Prove using mean value theorem. $|\arctan(\frac{a}{4})^{4}-\arctan(\frac{b}{4})^{4}|\leq \pi^{3}\cdot|a-b|$
I started with $f(x)=\arctan(\frac{x}{4})^{4}$.
The function is continuous on $[a,b]$ and differentiable in (a,b) 
So there exists a $c\in (a,b)$ such that  $f'(c)=\frac{\arctan(\frac{b}{4})^{4}-\arctan(\frac{a}{4})^{4}}{b-a}$. From this I have
$$|\frac{-1}{1+c^{2}}|= \frac{1}{1+c^{2}}=\frac{|\arctan(\frac{b}{4})^{4}-\arctan(\frac{a}{4})^{4}|}{|b-a|}$$
and $a<c<b$. How to get the inequality with $\frac{1}{1+c^{2}}$from it? 

Let $f(x) = x^4$, and suppose $|x| \le M$, then $|f'(x)| \le 4 M^3$ for $|x| \le M$.
Using the mean value theorem
we have $|f(x)-f(y)| \le 4 M^3 |x-y|$.
Let $M = {\pi \over 2}$ and note that $|\arctan x| \le M$ and
$|\arctan' x| \le 1$, hence
$|(\arctan x)^4 - (\arctan y)^4| \le {1 \over 2} \pi^3 |\arctan x -\arctan y | \le {1 \over 2} \pi^3 |x-y|$.
Now let $x = {a \over 4}, y = {b \over 4}$ to get.
$|(\arctan {a\over 4})^4 - (\arctan {b \over 4})^4|  \le {1 \over 2} \pi^3 |{a \over 4} -{b \over 4}| \le { \pi^3 \over 8} |x-y| $.
