If $f$ is $2 \pi$ periodic and $\int_{0}^{2 \pi} f(t) dt=0$ then $\int_{0}^{2 \pi} (f(t))^2 dt \le \int_{0}^{2 \pi} (f'(t))^2 dt$

Question

Given $f$ a real differentiable function, $2 \pi$ periodic such that $\int_{0}^{2 \pi} f(t) dt=0$ show that:

$\int_{0}^{2 \pi} (f(t))^2 dt \le \int_{0}^{2 \pi} (f'(t))^2 dt$. When does equality hold?