Let $f:[-\pi,\pi] \to \mathbb{R}$ be a piecewise smooth function with $\int_{-\pi}^{\pi}f(x)dx = 0$. Does anyone have ideas on how to apply Bessel's inequality to show that $\int_{-\pi}^{\pi} (f'(x))^2 dx \geq \int_{-\pi}^{\pi}(f(x))^2 dx$?
I've already shown that $\int_{-\pi}^{\pi}(f(x))^2 = a_{0}^{2}/2 + \sum_{n=1}^{\infty}(a_{n}^{2}+b_{n}^{2} )$
So, Bessel's Inequality states:
Let $f\in L_2[-L,L]$ then $$\frac{a_{0}^{2}}{2} + \sum_{n=1}^{\infty}(a_{n}^{2}+b_{n}^{2} ) \leq \frac{1}{L}\int_{-L}^{L}f^2(x)dx$$
Note that the Fourier Series of the first derivative of $f$ is $$f'(x) = \sum_{n=1}^{\infty}n\left[ a_n\cos(nx) + b_{n}\sin(nx)\right ]$$
So we essentially wish to show that $$\int_{-\pi}^{\pi}\left ( \sum_{n=1}^{\infty}n\left[ a_n\cos(nx) + b_{n}\sin(nx)\right ] \right )^2dx \geq \frac{a_{0}^{2}}{2} + \sum_{n=1}^{\infty}(a_{n}^{2}+b_{n}^{2} )$$
Does anyone know how to go from here? Should I expand and then try to simplify the integrals? However, won't we have trouble since it is not stated that we have uniform convergence?
As you noted correctly, taking a derivative amounts to multiplying the Fourier coefficients $a_n,b_n$ by $n$. The point in the inequality that you are after just comes from crudely estimating $(n a_n)^2\ge a_n^2$ and $(n b_n)^2\ge b_n^2$ for $n\ge 1$. Note also that $a_0=0$ by the mean zero assumption.
The argument goes as follows:
$$\frac{1}{2\pi}\int_{-\pi}^\pi (f'(x))^2 dx = \sum_{n=1}^\infty n^2 (a_n^2 + b_n^2) \ge \sum_{n=1}^\infty (a_n^2+b_n^2)=\frac1{2\pi} \int_{-\pi}^\pi (f(x))^2 dx$$
By the way, this is a very special case of the Poincaré inequality.