 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 giveApply the periodic Wirtinger inequality to each mean-zero coordinate: $\int |\gamma {|}^2\le (L/2\pi {)}^2\int |{\gamma }^{\prime }{|}^2$. Since unit speed gives the final integral $L$, it follows that $|A_{\mathrm{signed}}|\le L^2/(4\pi )$. For several boundary components, orient each with the domain on its left. Green's formula, followed by the triangle inequality, givesThis 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.