= Solution
The planar <planar isoperimetric inequality> is
$$
\boxed{4\pi A(\Omega)\leq l(\partial\Omega)^2},
$$
with equality precisely for a disc. Let the positively oriented boundary be parametrized by arclength as $r(s)=(x(s),y(s))$, $0\leq s\leq L$. Translate its mean to zero. Then $|r'|=1$, and <Green's theorem> gives $2A=\int_0^L(xy'-yx')ds$.
The periodic <Wirtinger inequality> says that for a real absolutely continuous $L$-periodic function $h$ of mean zero and square-integrable <derivative>,
$$
\int_0^Lh^2ds\leq\left(\frac L{2\pi}\right)^2\int_0^L(h')^2ds,
$$
with equality exactly for $h(s)=a\cos(2\pi s/L)+b\sin(2\pi s/L)$. Apply it to both coordinates and use <Cauchy-Schwarz inequality>:
$$
2A\leq\left(\int|r|^2ds\right)^{1/2}\left(\int|r'|^2ds\right)^{1/2}
\leq\frac L{2\pi}\int|r'|^2ds=\frac{L^2}{2\pi}.
$$
If equality holds, the Wirtinger equalities give $r=p\cos(2\pi s/L)+q\sin(2\pi s/L)$. The arclength condition forces $|p|=|q|=L/(2\pi)$ and $p\cdot q=0$, so the boundary is a circle. A circle attains equality, making the constant optimal.
\b[The inequality and equality characterization persist for finitely many corners.] A rectifiable piecewise smooth simple closed curve has a Lipschitz arclength parametrization, with $|r'|=1$ almost everywhere. Green's theorem and the absolutely continuous version of Wirtinger's inequality used above still apply. More generally one may approximate a rectifiable Jordan curve by smooth curves while preserving the limiting length and area. If the allowed isolated singularities give infinite length, the inequality is trivially satisfied in the extended sense.
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