Do nested convex bodies have increasing "Volume/Surface Area" ratios? Suppose we have two convex bodies $A$ and $B$, where $A \subseteq B$.  Is it always true that $\frac{\mathrm{Vol}(A)}{\mathrm{SurfaceArea}(A)}\leq \frac{\mathrm{Vol}(B)}{\mathrm{SurfaceArea}(B)}$?

It's true in all the examples I've tried, but I'm not sure how to prove the general case, or whether the general case is even true.
Counterexample:

Let $B$ be a unit square. It has area $1$ and perimeter $4$.

Truncate one corner of $B$ by chopping off an isosceles right triangle of base and height $\delta$; call this new pentagon $A$. It has area $1-\frac12\delta^2$ and perimeter $4 - 2\delta + \sqrt2\delta$.

We have $A\subseteq B$ but $$\frac{\text{Area}(A)}{\text{Perimeter}(A)} = \frac{1-\frac12\delta^2}{4 - 2\delta + \sqrt2\delta} &gt; \frac14 = \frac{\text{Area}(B)}{\text{Perimeter}(B)}$$
for any $\delta&lt;1-\frac1{\sqrt2}\approx0.293$.

I imagine a similar strategy works in higher dimensions as well.
One might note that $V=\int \vec r \cdot \mathrm{d}S$,  for each element of S.  If one figure contains a second figure, then this ratio increases, because in the containing figure $R \ge r$, and so the integral of R is greater than r.

This is a dot product, and it is worth noting that for polyhedra (bounded by flat planes), and cylinders, the dot product is basically based on the tangential insphere. 

So if we mean to find a counter-example, we might suppose that the outer figure has number of small inspheres, like a disphenoid tetrahedron, formed, eg by a square, with the diagonals of 1.4 (which would pull it in the fold).

The second figure would fit into this, but not venture out to the edges of the square.  The figure would need to exceed the insphere of the greater one, but there's plenty of room to fill out here.  

So it could be possible.