Consider the following problem (Exercise (2) in the end of Chapter 3 of Curves and Surfaces, 2nd Edition, by Montiel and Ros):
Let $S$ be a compact surface contained in a closed ball of radius $r > 0$. Prove that there exists at least one point $p \in S$ such that $K(p) \geq 1/r^2$ and $|H(p)| \geq 1/r$.
The book provides the following solution to this problem:
Let $a$ be the centre of the ball of radius $r$ in which $S$ is contained. By compactness, there is some $p \in S$ such that the square of the distance to $a$ attains a maximum. Then we know that $a$ is in the normal line of $S$ at $p$ and
$$
1 \leq \langle a – p, N(p) \rangle k_i(p), \quad i = 1, 2 \qquad (*)
$$
where $k_i$ are the principal curvatures. Keeping in mind that $\langle a – p, N(p) \rangle^2 = |a-p|^2 \leq r^2$, the proof follows bt adding or multiplying the inequalities $(*)$.
My question is:
Where does $(*)$ come from?
Best Answer
Let $f\colon S \to \Bbb R$ be given by $f(q) = \|q-a\|^2$. Then since $p$ is a maximum for $f$, you know that if $\alpha\colon (-\epsilon,\epsilon) \to S$ has $\alpha(0)=p$ and $\alpha'(0) = v$, then $$\frac{{\rm d}^2}{{\rm d}t^2}\bigg|_{t=0} f(\alpha(t)) \leq 0.$$Actually evaluating this gives $$(f\circ\alpha)'(t) = 2\langle \alpha'(t), \alpha(t)-a\rangle \implies (f\circ \alpha)''(0) = 2\langle \alpha''(0), p-a\rangle + 2\langle v,v\rangle \leq 0 $$This means that choosing $v$ to be an unit vector, you have $$1 \leq \langle \alpha''(0), a-p\rangle.$$But $p-a$ normalized is precisely $N(p)$, so only the normal part of $\alpha''(0)$ contributes to this product (recognize the definition of normal curvature), and you get $(\ast)$ by letting $v$ be the principal directions.