Suppose $S$ lies in the closed ball of radius $R$ centred at $c$. By <compactness>, the continuous function $q\mapsto|q-c|$ attains its maximum $r\leq R$ at some $p\in S$. The sphere of radius $r$ about $c$ is a <supporting sphere> tangent to $S$ at $p$.

Put $N=(p-c)/r$. For any unit vector $v\in T_pS$, choose a surface curve $\alpha$ with $\alpha(0)=p$ and $\alpha'(0)=v$. Since $|\alpha(t)-c|^2$ has a local maximum at zero,
$$
0\geq
\left.\frac{d^2}{dt^2}|\alpha(t)-c|^2\right|_{t=0}
=2+2r\langle\alpha''(0),N\rangle.
$$
The scalar $\langle\alpha''(0),N\rangle$ is the <normal curvature> in direction $v$, so every normal curvature is at most $-1/r$ with this choice of <unit normal>. Applying this to <principal direction>[principal directions] shows that both <principal curvature>[principal curvatures] satisfy $k_i\leq-1/r$. Therefore
$$
K(p)=k_1k_2\geq\frac1{r^2}\geq\frac1{R^2}>0.
$$
This proves both assertions, including the existence of an <elliptic point> on every compact regular surface.

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 25I / a / Solution

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An elliptic point of a surface is a point at which the <Gaussian curvature> is positive, equivalently where the two <principal curvature>[principal curvatures] have the same nonzero sign.

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 Incoming links: Elliptic point