Given that $f$ is $C^1$ on $\mathbb{R}$ with period $2\pi$ and $\int_{-\pi}^{\pi}f(t)dt =0$, show $$\int_{-\pi}^{\pi}[f'(t)]^2 dt \geq \int_{-\pi}^{\pi}[f(t)]^2 dt$$
I've tried using Bessel's inequality and Parseval's theorem, but I couldn't arrive at the inequality in question. I also tried integrating by part but couldn't get to a point where we could use $\int_{-\pi}^{\pi}f(t)dt =0$.
Thanks for the help.
Expand $f$ and $f'$ into Fourier series, $$f(x) = \sum_{n\in\mathbb{Z}}c_n e^{int},\quad f'(x) = \sum_{n\in\mathbb{Z}}b_n e^{int}$$ Here $b_n=inc_n$ by virtue of integration by parts (boundary terms cancel out): $$ b_n = \frac{1}{2\pi}\int_0^{2\pi} f'(t) e^{-int}\,dt = \frac{1}{2\pi}\int_0^{2\pi} f(t) in e^{-int}\,dt = in c_n $$ Since $c_0=0$, it follows that $|c_n|\le |b_n|$ for all $n$. The rest should be clear.