Integral inequality between function and derivative Let $f \in \mathcal{C}^1([0,1])$ such that  $f(0)=f(1)=0$.
I want to proof the following inequality:
$$\int_0^1 f'(x)^2 \,\mathrm{d}x \geq \pi^2 \int_0^1 f(x)^2 \,\mathrm{d}x $$
I proceeded by integration by parts to get 
$$\int_0^1 f''(x)f(x) \,\mathrm{d}x \geq \pi^2 \int_0^1 f(x)^2 \,\mathrm{d}x ,$$
but I have no idea what to do now.
Any help please ? 
 A: For each $n\in\mathbb{Z}_{> 0}$, define
$$a_n:=2\,\int_0^1\,\cos(2n\pi\,x)\,f(x)\,\text{d}x$$
and
$$b_n:=2\,\int_0^1\,\sin(2n\pi\,x)\,f(x)\,\text{d}x\,.$$
Furthermore, let
$$a_0:=\int_0^1\,f(x)\,\text{d}x\,.$$
Therefore,
$$f(x)=a_0+\sum_{n=1}^\infty\,a_n\,\cos(2n\pi\,x)+\sum_{n=1}^\infty\,b_n\,\sin(2n\pi\,x)\text{ for }x\in[0,1]\,.$$
Because $f(0)=0$ and $f(1)=0$, we must have
$$\sum_{n=1}^\infty\,a_{n}=-a_0\,.$$
We have
$$f'(x)=-\sum_{n=1}^\infty\,2n\pi\,a_n\,\sin(2n\pi\,x)+\sum_{n=1}^\infty\,2n\pi\,b_n\,\cos(2n\pi\,x)\text{ for }x\in[0,1]\,.$$
That is,
$$\int_0^1\,\big|f'(x)\big|^2\,\text{d}x=2\pi^2\, \sum_{n=1}^\infty\,n^2\,\big(|a_n|^2+|b_n|^2\big)$$
and
$$\int_0^1\,\big|f(x)\big|^2\,\text{d}x=|a_0|^2+\frac{1}{2}\,\sum_{n=1}^\infty\,\big(|a_n|^2+|b_n|^2\big)\,.$$
Thus,
$$\int_0^1\,\big|f'(x)\big|^2\,\text{d}x- \pi^2\,\int_0^1\,\big|f(x)\big|^2\,\text{d}x\geq \frac{\pi^2}{2}\,\left(\sum_{n=1}^\infty\,(4n^2-1)\,|a_{n}|^2-2\,|a_0|^2\right)\,.$$
Let 
$$w_n:=\frac{2}{4n^2-1}\text{ for }n=1,2,3,\ldots\,.$$
Note that $\sum\limits_{n=1}^\infty\,w_n=1$.
Then,
$$\int_0^1\,\big|f'(x)\big|^2\,\text{d}x- \pi^2\,\int_0^1\,\big|f(x)\big|^2\,\text{d}x\geq \frac{\pi^2}{2}\,\Biggl(\frac12\,\sum_{n=1}^\infty\,w_n\,\big((4n^2-1)|a_{n}|\big)^2-2\,|a_0|^2\Biggr)\,.$$
Since $\sum\limits_{n=1}^\infty\,w_n=1$, the Power-Mean Inequality implies that
$$\int_0^1\,\big|f'(x)\big|^2\,\text{d}x- \pi^2\,\int_0^1\,\big|f(x)\big|^2\,\text{d}x\geq \frac{\pi^2}{2}\,\Biggl(\frac12\,\left(\sum_{n=1}^\infty\,w_n\,(4n^2-1)|a_{n}|\right)^2-2\,|a_0|^2\Biggr)\,.$$
It follows immediately that
$$\int_0^1\,\big|f'(x)\big|^2\,\text{d}x- \pi^2\,\int_0^1\,\big|f(x)\big|^2\,\text{d}x\geq \frac{\pi^2}{2}\,\left(2\,\left(\sum_{n=1}^\infty\,|a_{n}|\right)^2-2\,|a_0|^2\right)\,.$$
Because $$\sum_{n=1}^\infty\,|a_{n}|\geq \left|\sum_{n=1}^\infty\,a_{n}\right|=|-a_0|=|a_0|\,,$$
we obtain
$$\int_0^1\,\big|f'(x)\big|^2\,\text{d}x- \pi^2\,\int_0^1\,\big|f(x)\big|^2\,\text{d}x\geq \frac{\pi^2}{2}\,\left(2-2\right)\,|a_0|^2=0\,.$$
Ergo,
$$\int_0^1\,\big|f'(x)\big|^2\,\text{d}x\geq \pi^2\,\int_0^1\,\big|f(x)\big|^2\,\text{d}x\,.$$
The equality becomes an equality if and only if there exist constant $A$ such that $$f(x)=A\,\left(\frac12-\sum_{n=1}^\infty\,\frac{\cos(2n\pi\,x)}{4n^2-1}\right)=a\,\sin(\pi\,x)$$ for $x\in[0,1]$, where $a:=\dfrac{4}{\pi}\,A$.
