A box contained within larger box has a smaller surface area than the larger box? Suppose we have a box (parallelepiped) A completely contained within another box B. Is the surface area of A nessecarily less than the surface area of the B?
Edit: note that the sides of A are not nessecarily parallel to the sides of B.
I happen to know that the answer is yes but the only solution I know of is very hand-wavy. 
 A: One form of Crofton's formula states that the area of a surface in $\mathbb R^3$ is proportional to the integral, over all planes, of the length of the intersection between the plane and the surface. (Reference: Stereology for Statisticians, equation 4.13)
Now, for any plane, its intersection with the inner box is a convex curve which is fully inside the intersection with the outer box. By another application of Crofton's formula in $\mathbb R^2$, the inner curve is shorter. So the inner box's integral is smaller. And so the inner box's area is smaller!
As achille hui noted in the comments, we don't even need the fact that the bodies are boxes. We just need them to be convex.
A: Oops, I have given a wrong proof before. Here is a corrected version:
Let $K$ be any convex body in $\mathbb{R}^3$, i.e. bounded convex subset of $\mathbb{R}^3$ which is convex (with non-empty interior). 
Let $u$ be any direction represented as a unit vector in $S^2$. 
Consider the orthogonal projection of $K$ onto a plane with normal vector $u$
and let $f(K,u)$ be the area of the projected image.
Cauchy surface area formula states that the average of $f(K,u)$ over $u$ is equal to $\frac14$ of the surface area of $K$. If we parametrize $u$ by spherical polar coordinates 
$$[0,\pi] \times [0,2\pi] \ni (\theta,\phi) \quad\mapsto\quad
u = (\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta) \in S^2$$
This is equivalent to the integral identity:
$$\frac{1}{4\pi}\int_0^{2\pi}\int_0^{\pi} f(K,u(\theta,\phi)) \sin\theta d\theta d\phi = \frac14 \verb/Area/(K)$$
When $K$ is a box, it is easy to verify this yourself. After a little algebra,
everything comes down to evaluation of a simple integral:
$$
\frac{1}{4\pi}\int_0^{2\pi}\int_0^{\pi} \max(\cos\theta,0)\sin\theta d\theta d\phi
= \frac14$$
Back to our problem, if we are given two boxes $A, B$ such that $A \subset B$, then for all unit vectors $u$, we have $f(A,u) \le f(B,u)$. Taking averages over
$u$ immediately give us $\verb/Area/(A) \le \verb/Area/(B)$.
