There is a theorem:
Let $f$ be a continuously differentiable, $2\pi$-periodic function. Given $\int_{-\pi}^{\pi} f(x) dx = 0$, I need to prove that
$$||f|| \le \frac{\pi}{2} \cdot ||f'||.$$
Where norm is maximum of modulus (not sqrt from integral $f^2$)
This inequality between norm of function and it's derivative was considered by D.G. Northcott in 1939. There is link to oxford journal, where he's article was placed. You can see, that he has a theorem about nth derivative of function. There is formula of coefficients for any n. You can see, that it is the same formula for the first derivative.
The problem is that access to full article (4 pages) is not free and I can't find proof of theorem anywhere. And I have no idea, how to prove inequality for $n=1$. I think it is not trivial: from Northcott's formula we get $\pi/2$ as $$\frac{4}{\pi} \cdot \sum_{k=1}^{\infty} \frac{1}{(2k-1)^2} = \frac{4}{\pi} \cdot (\frac{\pi^2}{6}-\frac{\pi^2}{24}) = \frac{4}{\pi} \cdot \frac{\pi^2}{8} = \frac{\pi}{2}.$$
So, we get $\pi^2$ magically from sum and it helps us to get $\pi$ in numerator. For second derivate we even get smth like $\zeta(3)/\pi$. So, there is no idea what to start from.
I hope somebody can help with proof. Or you never know, somebody has access to Oxford Journal :)