Understanding an example for "minimal surface doesn't imply least area" I can't understand two things regarding the following example: 

1- Why the minimal surface $S$ will not minimize the area among all surfaces with boundary the two circles if $S_0 < S$? I don't understand this phrase at all and I don't know why the inequality must hold?
2- What is the meaning of the last paragraph and how to show that it holds (a proof)?
A clear simple explanation would be much appreciated.
PS - Source is the book Elementary Differential Geometry by A. N. Pressley.
 A: The exposition is somewhat unclear because they use $a$ twice in two different contexts.  For the sake of this discussion, consider only the function $$x = \cosh z, \quad -\infty < z < \infty,$$ from which the minimal surface is formed by the surface of revolution $r(z) = \cosh z$ rotated about the $z$-axis.  All subsequent use of the letter $a$ is in the context of some positive constant that can be arbitrarily chosen.
The whole catenoid, of course, is an infinite surface.  But if we select a portion of the catenoid satisfying $|z| \le a$, then we get what might seem to be a surface $S$ whose area minimizes the surface area of any surface that "spans" the boundary.  But this is not true:  the surface area of the catenoid between the planes $z = a$ and $z = -a$ is easily found even by elementary methods, as it is a surface of revolution defined by the function $r(z)$:  $$|S| =  \int_{z=-a}^a 2 \pi r(z) \sqrt{1 + \left(\frac{dr}{dz}\right)^2} \, dz = 4\pi \int_{z=0}^a \cosh^2 z \, dz = 4\pi \left[\frac{z}{2} + \frac{\sinh 2z}{4}\right]_{z=0}^a =  \pi (2a + \sinh 2a).$$  This is equivalent to the area described in the text.  But the combined area of the two disks is just $$|S_0| = 2(\pi \cosh^2 a),$$ because their common radius is $r(a) = r(-a) = \cosh a$.  Now we want to see which quantity is smaller for a given $a > 0$:  to this end, we set  $$2\pi \cosh^2 a < \pi(2a + \sinh 2a),$$ and expanding this into exponential form, and simplifying, we get $$1+e^{-2a} < 2a,$$ as claimed.  This means that whenever $a$ satisfies this inequality, the disks will have less area than the catenoid.  Our intuition should tell us that this happens for "sufficiently large" $a$:  indeed, numerically solving this inequality gives $a > a_0 \approx 0.639232\ldots$.  This is because when $a$ gets large, the catenoid becomes much more "disk-like" at the boundary, whereas if $a$ is small, the catenoid is "cylinder-like".  But simply having two disks is going to clearly win out over two deformed funnel-like shapes that have been made to join at their centers.

