Inequality - find the maximum possible constant here is the question
Find the maximum possible constant so it satisfies the following:
$$\int_0^{2π} (u'(x))^2  dx \geq c\int_0^{2π} (u(x))^2  dx $$
with:
$$u\in C^1[0,2π] ,u(0)=u(2π)=0 $$
*My thought was that I should use the Cauchy–Schwarz inequality like this $$(\int_0^{2π} udx)^2 \leq \int_0^{2π}dx\int_0^{2π}u^2dx=2π \int_0^{2π}u^2dx  $$
and $$U(x)=\int_0^{y}u'(y)dy$$ so $$(\int_0^{y}u'(y)dy)^2\leq \int_0^{x}dy\int_0^{x}u^2dy=x\int_0^{x}u^2dy$$ , but I am not sure what to do next really, or if I am going anywhere, thanks.
 A: Note that $\{\frac{1}{\sqrt\pi}\sin(\frac{nx}{2})\}_{n=1}^\infty$ is dense in $A=\{u\in C^1[0,2\pi]: \ u(0)=u(2\pi)=0\}$. Also note
\begin{eqnarray}
&\int_0^{2\pi}\sin^2(\frac{nx}{2})dx=\int_0^{2\pi}\cos^2(\frac{nx}{2})dx=\pi, \\
&\int_0^{2\pi}\sin(\frac{mx}{2})\sin(\frac{nx}{2})dx=\int_0^{2\pi}\cos(\frac{mx}{2})\cos(\frac{nx}{2})dx=0\text{ for }m\neq n.
\end{eqnarray}
Let $u\in A$. Then $u$ can be written as
$$ u=\sum_{n=1}^\infty\frac{a_n}{\sqrt \pi}\sin(\frac{nx}{2}) $$
and hence
$$ u'=\sum_{n=1}^\infty\frac{n}{2\sqrt \pi}\cos(\frac{nx}{2}). $$
where $a_n, n=1,2,3,\cdots$ are reals.
So
$$ \int_0^{2\pi}u^2dx=\sum_{n=1}^\infty a_n^2, \int_0^{2\pi}(u')^2dx=\sum_{n=1}^\infty \frac{n^2}{4}a_n^2. $$
If 
$$ \int_0^{2\pi}(u')^2dx\ge C\int_0^{2\pi}u^2dx$$
then one must have 
$$ \frac{n^2}{4}a_n^2\ge Ca_n^2, n=1,2,\cdots$$
from which one can see that the maximum possible constant for $C$ is $C=\frac14$. It is easy to see that if $u=\frac{1}{\sqrt \pi}\sin(\frac{x}{2})$, the equal sign holds for $C=\frac14$.
