TITLE: Minimum value of $ \displaystyle \frac{\int_{0}^{1}{[f'(x)]^2 \ dx}}{\int_{0}^{1}{[f(x)]^2 \ dx}} $
QUESTION [1 upvotes]: Let there be a function $f(x)$ double differentiable in $[0,1]$ and it is given that $f(0) = f(1) = 0$ while $f(x)$ is not identically zero. So find the smallest possible value of 
$$ \displaystyle \frac{\int_{0}^{1}{[f'(x)]^2 \ dx}}{\int_{0}^{1}{[f(x)]^2 \ dx}} $$
I have tried using IBP which seems inevitable to the first sight but of not much significance, u-substitution too does not make it quite clear. 
How to do such kinds of problems anyway? It doesn't seems to me that it requires any advanced stuff.

REPLY [3 votes]: When $f(x)=\sin(\pi x)$, we get the value $\pi^2$. To show that this is the minimum, consider any function $f$ with $\|f\|_2=1$ that has a Fourier expansion. Note that only $\sin(\pi nx)$ terms appear, due to the boundary conditions. Applying Parseval's and evaluating the Rayleigh-Ritz quotient yields a weighted average of the eigenvalues
$$
\pi^2, 4\pi^2,\ldots,
$$
with weights given by the Fourier coefficients of the expansion.
Thus the minimum value is $\pi^2$.
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