= Solution
The planar <planar isoperimetric inequality> is $4\pi|\Omega|\le|\partial\Omega|^2$, with equality exactly for a disk. For one boundary component of length $L$, take its unit-speed periodic parametrization $\gamma=(x,y)$ and subtract its mean position. Green's area formula and the <Cauchy-Schwarz inequality> give
$$
|A_{\rm signed}|=\frac12\left|\int_0^L(xy'-yx')ds\right|\le\frac12\left(\int_0^L|\gamma|^2ds\right)^{1/2}\left(\int_0^L|\gamma'|^2ds\right)^{1/2}.
$$
Apply the periodic <Wirtinger inequality> to each mean-zero coordinate: $\int|\gamma|^2\le(L/2\pi)^2\int|\gamma'|^2$. Since unit speed gives the final integral $L$, it follows that $|A_{\rm signed}|\le L^2/(4\pi)$. For several boundary components, orient each with the domain on its left. Green's formula, followed by the <triangle inequality>, gives
$$
|\Omega|\le\frac1{4\pi}\sum_jL_j^2\le\frac1{4\pi}\left(\sum_jL_j\right)^2.
$$
This proves the stated inequality, including disconnected domains or holes. Equality permits only one component; equality in the two integral inequalities makes its coordinates first sine-cosine harmonics with the corresponding orthogonal rotation relation. Unit speed then makes the curve a circle, so its domain is a disk. Conversely disks attain equality.
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