Good morning Nickos,

We are basing the whole fact on Fourier series. Since all Dirichlet's conditions are met ,then \( f \) can be expanded into a Fourier series. Therefore we can write it as:

$$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left [ a_n \sin nx+b_n \cos nx \right ]$$

However, since the integral of \( f \) vanishes , so does \( a_0 \). Applying Pasheval's identity we get:

\( \bullet \;\;\; \displaystyle \int_{0}^{2\pi}f^2(x)\,{\rm d}x=\sum_{n=1}^{\infty}\left ( a_n^2+b_n^2 \right ) \) and

\( \bullet \;\;\; \displaystyle \int_{0}^{2\pi}\left ( f'(x) \right )^2\,{\rm d}x=\sum_{n=1}^{\infty}n^2\left ( a_n^2+b_n^2 \right ) \)

Finally , since all summands are positive we get the desired inequality and the exercise comes to and end.