Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 2 / 25H / c / 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 $k(s_0)\le 2\pi /L$. The planar isoperimetric inequality gives $L^2\ge 4\pi A(K)$, henceFor 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.