= Solution
The <periodic Wirtinger inequality> says that if $f$ is continuously differentiable, $L$-periodic, and has mean zero, then
$$
\int_0^L f(s)^2\,ds
\leq \frac{L^2}{4\pi^2}\int_0^L f'(s)^2\,ds.
$$
Equality holds exactly for a linear combination of $\cos(2\pi s/L)$ and $\sin(2\pi s/L)$.
First suppose $\partial\Omega$ is one positively oriented simple closed curve, parametrized by <arc length> as $\gamma(s)=(x(s),y(s))$ for $0\leq s\leq L$. Translating the origin, which changes neither length nor area, makes both $x$ and $y$ have mean zero. <Green theorem> and the <Cauchy-Schwarz inequality> give
$$
\begin{aligned}
2\operatorname{Area}(\Omega)
&=\int_0^L(xy'-yx')\,ds\\
&\leq
\left(\int_0^L(x^2+y^2)\,ds\right)^{1/2}
\left(\int_0^L((x')^2+(y')^2)\,ds\right)^{1/2}.
\end{aligned}
$$
Applying Wirtinger's inequality to both coordinate functions and using the unit-speed identity $(x')^2+(y')^2=1$ yields
$$
2\operatorname{Area}(\Omega)
\leq\frac{L}{2\pi}\,L,
$$
and therefore the <planar isoperimetric inequality>
$$
\boxed{4\pi\operatorname{Area}(\Omega)\leq L^2}.
$$
For several boundary components, apply the simple-curve result to the relevant enclosed regions and use $\sum_jL_j^2\leq(\sum_jL_j)^2$; holes only decrease the area.
Equality in both inequalities forces
$$
(x,y)=A\cos(2\pi s/L)+B\sin(2\pi s/L)
$$
for constant vectors $A,B$, with the unit-speed and Cauchy-Schwarz equality conditions making $A$ and $B$ perpendicular and equally long. Thus the boundary is a circle. Conversely, a circular domain has area $\pi R^2$ and perimeter $2\pi R$, so equality holds.
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