The following is Exercise (18) at the end of Chapter 3 in Curves and Surfaces, 2nd edition, by Montiel and Ros:
Let $S$ be the graph over a disc in $\Bbb{R}^2$ with radius $r > 0$. If $S$ has mean curvature $H \geq a$ at each point for some real number $a$, show that $ar \leq 1$.
The execise is solved at the end of the chapter, but I did not quite get it. The solution provided is as follows:
If $a \leq 0$ the result is obvious. Assume $a > 0$. Take $0 < R < r$ and let $\Bbb{S}_R^2$ be a sphere with radius $R$ that has constant mean curvature $1/R$ with respect to its inner normal. Assume that $S$ is a graph over a disc centered at the origin of the plane $z = 0$. Drop the sphere $\Bbb{S}_R^2$ onto $S$ in the two possible ways and obtain, using exercise (20) at the end of Chapter 2, two points of tangency between $S$ and $\Bbb{S}_R^2$. At least at one of them, say $p$, the normal of $S$ coincides with the inner normal of $\Bbb{S}_R^2$ Now use exercise (17) to obtain $$ a \leq H(p) \leq 1/R. $$ Then $aR \leq 1$ for all $R < r$.
My questions are:
What do the authors mean by "Drop the sphere $\Bbb{S}_R^2$ onto $S$ in the two possible ways"? Which ways are these?
Below are the statements of Exercises (20) of Chapter 2 and (17):
(20): Suppose two surfaces $S_1$ and $S_2$ cut it other transversely at a point $p \in T_pS_1 \cap T_pS_2$. Prove that if $t \in \Bbb{R}$ is small enough then $\phi_t(S_1) \cap S_2 \neq \emptyset$ where $\phi_t$ is the translation of $\Bbb{R}^3$ given by $\phi_t(p) = p + ta$ where $a \in \Bbb{R}^3$ with $|a| = 1$.
(17) Let $S_1$ and $S_2$ be two orientable surfaces that are tangent in a common point $p \in S_1 \cap S_2$ and let $N_1$ and $N_2$ be the corresponding Gauss maps such that $N_1(p) = N_2(p)$. We express $S_1$ and $S_2$ near the point $p$ as two graphs of functions $f_1$ and $f_2$ defined on the common tangent plane. We say that $S_1$ is over $S_2$ at $p$ if there exists a neighborhood of the origin of this plane where $f_1 \geq f_2$. Prove that if $S_1$ is over $S_2$ at $p$ then $\sigma_1(v, v) \geq \sigma_2(v, v)$ for each $v \in T_pS_1 = T_pS_2$. In particular, $H_1(p) \geq H_2(p)$. Show that if $\sigma_1(v, v) > \sigma_2(v, v)$ for each $v \neq 0$ then $S_1$ is over $S_2$ at $p$.
$\sigma$ denotes the second fundamental form.
Thanks in advance and kind regards.