= Solution
For the positively oriented simple closed boundary, the <turning tangent theorem> gives $\int k(s)\,ds=2\pi$. If $L$ is its length, continuity gives a point with $k(s_0)\leq2\pi/L$. The <planar isoperimetric inequality> gives $L^2\geq4\pi A(K)$, hence
$$
\boxed{k(s_0)\leq\frac{2\pi}{L}\leq\sqrt{\frac\pi{A(K)}}.}
$$
For the reversed orientation the <curvature> <integral> is $-2\pi$, so a point of nonpositive <curvature> already proves the stated positive upper bound. The region-bounding interpretation is the ordinary Jordan-curve one; self-intersections would require specifying a different notion of enclosed area.
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