Known proofs of Wirtinger's Inequality? I am looking for proofs of the (Poincare-) Wirtinger inequality which states that if $f:[0,\pi]\to \mathbb{C}$ is $C^1$ and $f(0)=f(\pi)=0$ then
\begin{equation}
\int_0^\pi |f(t)|^2 dt \leq \int_0^\pi |f'(t)|^2 dt.
\end{equation}
See link.

The proof that I know starts by proving that if
$$ \int_0^{2\pi} F(t) dt =0 $$
then
$$
\int_0^{2\pi} |F(t)|^2 dt \leq \int_0^{2\pi} |F'(t)|^2 dt.
$$
using Parseval's identity. From this, one proves the desired inequality for $f$ on $[0,\pi]$ by "extending" $f$ to an odd $C^1$ function on $[-\pi,\pi]$.

Are there other proofs? (Straightforward or otherwise...)
The following proof is found in section 7.7 of Hardy-Littlewood-Polya Inequalities, it is motivated by Hilbert's investigations into calculus of variations, especially Hilbert's method of invariant integrals. 

Consider the expression
$$ (y'^2 - y^2) - (y' - y\cot x)^2 = -(1+ \cot^2 x) y^2 + 2y y' \cot x $$
So
$$ \left[(y'^2 - y^2) - (y' - y\cot x)^2 \right]\mathrm{d}x = -(\csc^2 x)y^2 \mathrm{d}x + 2y \mathrm{d}y \cot x = \mathrm{d} ( y^2 \cot x )$$

Now, since $y' \in L^2$, we have that
$$ y^2(x) = \left(\int_0^x y'(s) \mathrm{d}s\right)^2 \leq \int_0^x y'(s)^2 \mathrm{d}s \int_0^x 1\mathrm{d}s \leq x \int_0^x y'(s)^2 \mathrm{d}s $$
So we have that
$$ \frac{y^2(x)}{x} = o(1) $$
and hence $y = o(\sqrt{x})$. Similarly we have that $y^2$ approaches 0 superlinearly at $\pi$. This implies that $\lim_{x\to \{0,\pi\}} y^2 \cot x = 0$. Hence the exact integral

$$ \int_0^\pi (y'^2 - y^2) - (y' - y\cot x)^2 \mathrm{d}x = \int \mathrm{d}\left( y^2 \cot x\right) = 0 - 0 = 0 $$

Therefore we have

$$ \int_0^\pi (y'^2 - y^2) \mathrm{d}x = \int_0^\pi (y' - y\cot x)^2 \mathrm{d}x \geq 0 $$

with equality only if 

$$ y' = y \cot x $$

which is when $y = k \sin x$.
You can find different proof in the book of B. Dacorogna 'Introduction to the calculus of variations'.