I doubt that I have the answer, but I can give a value (an upper bound)
which is probably rather close to the minimum ratio. For those who wish
to work on this problem before seeing what I've done so far,
First, note that, just as in "The Free Plot", many people will tend to
jump to wrong conclusions, thinking that either the given cube itself
or the sphere inscribed therein will have the minimum ratio of surface area
to volume. In fact, the cube and the sphere have the same ratio, 6/e, where
e denotes the edge length of the cube. Henceforth, let's assume that we are
dealing with a cube of unit edge length, so the surface area to volume
ratio for both the cube and the sphere is 6. But this is certainly not the
minimum ratio obtainable. A solid produced by rounding the edges and
corners of the cube somewhat can have a smaller ratio.
Proceeding by analogy with the solution to "The Free Plot", I considered
a solid formed from the given cube by rounding all of its edges, leaving
them "quarter round" with radius r. Thus, any orthogonal projection
of the solid onto an original cube face would resemble the solution to
"The Free Plot" (although the radius r of the quarter circles would not
necessarily be 1/(2 + Sqrt(pi)), of course). Also consider the parts of
the solid contained within small cubes of edge length r nestled in the
corners of the original cube. These eight parts could be assembled to form
a tricylinder (a Steinmetz solid, the intersection of three cylinders). It
is only on these parts that the solid has edges or corners. They are rather
blunt, but surely the fact that the solid has them at all tells us that a
still smaller ratio of surface area to volume can be obtained, by using a
smooth solid. [Unfortunately, I don't know precisely what that smooth solid
should be. It might be interesting to investigate solids given by
|x|^p + |y|^p +|z|^p <= 1/2^p for different values of p.]
For the type of solid I investigated, obtained by "quarter rounding" of
the cube of unit edge length,
surface area A = 6(1-2r)^2 + 6(pi)r(1-2r) + 24(2-Sqrt(2))r^2 and
volume V = (1-2r)^3 + 6r(1-2r)^2+ 3(pi)r^2(1-2r) + 8(2-Sqrt(2))r^3.
Minimizing A/V for such a solid is straightforward, leading to the
solution of a quartic equation. Rather than giving a very messy precise
expression for the value of r at which the minimum is attained, let me
just say that it is approximately r = 0.2376425, at which A/V is
approximately 5.45525 .
As noted previously, surely this is not the best possible, due to the
corners and edges of this type of solid. I have no idea whether finding
the solid giving minimum A/V will be difficult or not, but it seems like
an interesting problem.
Happy New Year!
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may be apposite.