= Supporting-sphere curvature bound
{title2=$K(p)\geq R^{-2}$}
Let a nonempty compact regular <surface> without boundary lie in a closed <Euclidean ball> of radius $R$. Then it has a <point> $p$ at which its <Gaussian curvature> satisfies $K(p)\geq R^{-2}$.
Maximize the squared distance to the center $c$. Put $\rho=|X(p)-c|$, so $0<\rho\leq R$. Its first <derivative> makes the radial <vector> normal to the <surface>; choose the <unit normal> $N=(X(p)-c)/\rho$. For a unit <tangent vector> $v$, the second <derivative> at this maximum gives
$$
0\geq\tfrac12\operatorname{Hess}_M|X-c|^2(v,v)=1+\rho\,II(v,v).
$$
Both <principal curvatures> are therefore at most $-1/\rho$. Their product is at least $1/\rho^2\geq1/R^2$. Reversing the <unit normal> reverses both signs and leaves the <Gaussian curvature> unchanged.
The no-boundary hypothesis is necessary: a flat planar <closed disk> lies inside a ball but has zero <Gaussian curvature> everywhere. Its farthest <points> occur on the boundary, where the interior maximum argument does not apply.
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