= Solution
For a regular simple closed plane curve of length $L$ enclosing area $A$, the <planar isoperimetric inequality> is \b[$L^2\geq4\pi A$], with equality precisely for a circle.
The periodic <Wirtinger inequality> states that an absolutely continuous $L$-periodic function $h$ of mean zero with square-integrable derivative satisfies
$$
\int_0^L h^2\,ds\leq\left(\frac L{2\pi}\right)^2\int_0^L(h')^2\,ds,
$$
with equality only for a linear combination of $\sin(2\pi s/L)$ and $\cos(2\pi s/L)$. Parametrize the curve by <arc length> and translate its position vector $r=(x,y)$ so that both coordinate means vanish. Then $\int|r|^2\leq L^3/(4\pi^2)$, since $|r'|=1$. <Green's theorem> and the <Cauchy-Schwarz inequality> give
$$
A=\frac12\left|\int_0^L(xy'-yx')\,ds\right|
\leq\frac12\left(\int|r|^2\right)^{1/2}\left(\int|r'|^2\right)^{1/2}
\leq\frac{L^2}{4\pi}.
$$
Equality requires a first-harmonic parametrization and equality in Cauchy-Schwarz; its radius vector is a constant multiple of the rotated unit tangent. It is therefore a circle. Conversely, a circle attains equality.
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