Given any compact surface in $\mathbb{R}^3$, there is a point with positive Gaussian Curvature

So if we have a compact surface $$S \subset \mathbb{R}^3$$ then the function $$f:S\rightarrow \mathbb{R}_{\geq 0 },\ x \mapsto ||x||$$ has a maximum value say $$R >0$$ at $$p \in S$$. Then we must have the surface $$S$$ fully enclosed by the sphere of radius $$R$$ as otherwise $$f(p)$$ is not the maximum value. This means that the curvature $$\kappa$$ at $$p$$ must be at least $$1/R \$$ for any curve on $$S$$ passing through $$p.$$

This next part is what I don't understand

So every normal curvature is at least $$1/R$$ so $$K(p)\geq 1/R^2 >0$$.

Why must every normal curvature be at least $$1/R$$ and why should that then mean $$K(p)\geq 1/R^2$$ ? For the final part I suspect it is because the Gaussian curvature is the product of the principal curvatures, but why should they be at least $$1/R$$ ?

• For each tangent vector $v$ at $p$ there is a curve on $S$ in that direction whose curvature is all in the normal direction, namely the intersection of $S$ with the plane spanned by $v$ and the normal to $S$ at $p$. Apr 20 at 18:26

Because if there is a curve in $$S$$ through $$p$$ with less curvature, it wouldn't be able to stay inside the sphere centered at the origin with radius $$R$$. And it has to stay inside that sphere, as that sphere contains all of $$S$$.
• I understand that the curvature of any curve should be no less than $1/R$ but why should the this be true for the normal curvature, where $\kappa ^2 = \kappa _n^2 +\kappa _g^2$?
• @Anon It's true for any curve, including geodesics ($\kappa_G=0$). Apr 20 at 17:03
• But if $1/R^2 \leq \kappa ^2 = \kappa _n^2 +\kappa _g^2$ why does this necessarily mean $\kappa _n \geq 1/R$ ?
• @Anon Because for any curve in $S$ through $p$ there is a geodesic that's tangent to it and has the same $\kappa_n$. That's what $\kappa_g$ is for: measure how far the curve is from being a geodesic (that's what the $g$ stands for). Apr 20 at 20:45