How can I prove, that for any real $p$ and $q$ such that $|p-q|<{{\pi}\over{2}}$
The inequality
$$4 \cdot \pi^3 \cdot |\tan(p)-\tan(q)| \ge |p^4-q^4|$$
is true
I am dividing by $p-q$ to get: $$4 \cdot \pi^3 \cdot \left|{\tan(p)-\tan(q)\over p-q}\right| \ge |(p+q) \cdot (p^2+q^2)|$$ I can assume that $p>q$ without loss of generality. Now using the mean value theorem, there must exist $c \in (p,q)$ such that $\tan(c) = \left|{\tan(p)-\tan(q)\over p-q}\right|$
So finally i'll have to prove this: $$4 \cdot \pi^3 \cdot |\tan(c)| \ge |(p+q) \cdot (p^2+q^2)|$$
Proving this for $|p-q| \ge {\pi \over 2} $is really simple, but I can't handle the other case. Can you please spare some help?
Edit
Thanks to @Thomas Anders I have noticed that my equation is wrong. Ideed, I didn't take the derivative. The right form of the equation is:
$$4 { \pi^3 \over|\cos^2(c)|} \ge |(p+q) \cdot (p^2+q^2)|$$
Now, because values of the cosine are in range from 0 to 1, the minimum value of the left hand side is $4*pi^2$
Now, let's take care of the right hand side. I have assumed, that $p>q$, so I can approximate the left hand side like this: $$|(p+q) \cdot (p^2+q^2)| \lt |(2q+{\pi \over 2})\cdot q^2\cdot({\pi^2 \over 4}+1)|$$
Now I differentiate the left hand side, and find out that it has a local maximum in $x={-\pi \over 6}$, and for any $q$ greater than zero it is growing. The right hand side also has a limit of infinity when $q$ goes to infinity. So, now I have proven the unequality for any $q$ lesser than 0, but how can I do this for $q$ greater than zero?