Exercice on periodic function

Question

Let $f$ be a periodic function, $\mathcal{C}^1$ on $\mathbb{R}$ such that: $$\displaystyle\int_0^{2 \pi} f(t) \, dt = 0$$

$$f(2 \pi) = f(0)$$

Prove that $$\forall t \in [0,2 \pi]: \int_0^{2 \pi} |f(t)|^2 dt \leq \int_0^{2 \pi} |f'(t)|^2 dt$$

How can we prove this please. I don't have any idea.

Answer 1

Hint: Expand $f$ in Fourier series.