= Solution
For the <Hurwitz proof of the planar isoperimetric inequality>, take a positively oriented regular simple closed curve of length $L$ and enclosed area $A$. Write its complex position as $z(t)=x(t)+iy(t)$, with $0\leq t\leq2\pi$ proportional to <arc length>. Then $|z'(t)|=L/(2\pi)$. Translate the curve to make its mean position zero, and write its <Fourier coefficients> as $c_n$, with $c_0=0$.
<Green's theorem> gives the signed area, and <Parseval's identity> computes it:
$$
A=\frac12\int_0^{2\pi}(xy'-yx')\,dt
=\frac12\operatorname{Im}\int_0^{2\pi}\overline z\,z'\,dt
=\pi\sum_{n\in\mathbb Z}n|c_n|^2.
$$
Periodic <integration by parts> and <Parseval's identity> applied to $z'$ give
$$
\sum_n n^2|c_n|^2=\frac1{2\pi}\int_0^{2\pi}|z'|^2\,dt
=\frac{L^2}{4\pi^2}.
$$
Since $n\leq n^2$ for every integer $n$,
$$
\boxed{A\leq\pi\sum_n n^2|c_n|^2
=\frac{L^2}{4\pi},\qquad L^2\geq4\pi A.}
$$
The sums converge absolutely: $\sum|n||c_n|^2\leq(\sum|c_n|^2)^{1/2}(\sum n^2|c_n|^2)^{1/2}$. Equality forces $c_n=0$ unless $n=0$ or $1$; after the mean translation, $z(t)=c_1e^{it}$ is a circle. Conversely a circle attains equality. Thus \b[circles uniquely attain equality, up to translation and orientation].
The same proof applies to a rectifiable simple closed curve using its Lipschitz <arc length> parametrization. Its derivative exists almost everywhere and belongs to $L^2$; periodic <mollification> converges to the curve in the function and derivative $L^2$ norms. This justifies the derivative coefficient identity, <Parseval's identity> and area integral by approximation. Reversing orientation, if necessary, makes the enclosed area positive. The printed name “Hurewitz” is read as Hurwitz.
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