= Solution
The preceding parts show that an area maximizer among curves of fixed length is a plane section of the sphere, hence a circle. If its smaller cap has angular radius $0\leq\theta\leq\pi/2$, then
$$
L=2\pi\sin\theta,\qquad
A=2\pi(1-\cos\theta).
$$
For this circle,
$$
A(4\pi-A)
=4\pi^2(1-\cos^2\theta)
=4\pi^2\sin^2\theta=L^2.
$$
\b[No curve of the same length encloses more area than the maximizing circle.] Equivalently every curve enclosing the smaller area $A$ satisfies the <spherical isoperimetric inequality>
$$
\boxed{L^2\geq A(4\pi-A)},
$$
with equality precisely for circles.
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