Showing an inequality on a definite integral 
Claim: For every $y \in C^1[0,1]$ with $y(0) = y(1) = 0$, the inequality
  $$\int_0^1y'(x)^2dx \geqslant\pi^2\int_0^1y(x)^2dx$$
  holds.

How may this be shown? The direction is giving me trouble as most inequality proofs go in the opposite direction.
 A: Some details in my answer may require additional mathematical validation, but I think the flow of the proof should be as follows.
The function $y$ could be continued on the interval $[-1,0]$ so that the resulting function will be odd. For this function $(y\in C^1[0,1])$ we have pointwise convergent Fourrier series. The function is odd, so we have only sinus functions
$$
y(x) = \sum_{n=1}^\infty b_n\sin\left(\pi nx\right).$$
If we take the derivative
$$
y^\prime(x) = \sum_{n=1}^\infty \pi nb_n\cos\left(\pi nx\right).$$
According to Parseval's identity
$$\int_0^1y'(x)^2dx = \sum_{n=1}^\infty (\pi nb_n)^2 = \pi^2\sum_{n=1}^\infty ( nb_n)^2\geqslant \pi^2\sum_{n=1}^\infty b_n^2 = \pi^2\int_0^1y(x)^2dx$$ 
Edit. 
As JeanMarie and uniquesolution have stated in the comments to the post, we should justify our term-by-term differentiation by the condition that $y\in C^1[0,1]$.
A: This is called Wirtinger's inequality. The ordinary proof uses Fourier series, but a nice proof of this can by done by hubristically assuming that we can write
$$ \int_0^1 (y'^2-\pi^2y^2) \, dx = \int_0^1 (y'-\psi y)^2 \, dx $$
for some function $\psi$. When will this be the case? Expanding out,
$$ \int_0^1 (y'-\psi y)^2 \, dx = \int_0^1 (y'^2-2\psi yy'+\psi^2 y^2) \, dx, $$
and integrating the middle term by parts,
$$ \int_0^1 -2\psi yy' \, dx = [-\psi y^2]_0^1 + \int_0^1 \psi' y^2 \, dx.  $$
Hence we can make this identification if $\psi'-\psi^2 = -\pi^2$, which has solutions
$$ \psi = \pi\cot{(\pi x+c)}. $$
We also need $\psi y^2$ to vanish at the endpoints, and $\psi$ to be continuous elsewhere. Since $y\in C^1[0,1]$, $y^2 = O(x^2)$ at the endpoints, so we can get away with choosing $c=0$, and we conclude that
$$ \int_0^1 (y'^2-\pi^2y^2) \, dx = \int_0^1 (y'-y\pi\cot{\pi x})^2 \, dx \geqslant 0, $$
with equality if and only if $y'-y\pi\cot{\pi x} = 0$, or $y=k\sin{\pi x}$.
(The above can be found in more detail in the classic Hardy, Littlewood and Pólya, Inequalities, p. 184f.)
