Suppose a box A is contained in a box B. Is the total volume of the k-faces of A at most the total volume of the total k-faces of B? This question inspired by question 2 from here.
Say we have two $n$-dimensional boxes (not necessarily axis-aligned, otherwise the problem is trivial) $A$ and $B$, with $A$ contained in $B$.  Choose $1\le k\le n$.  Is the total volume of the $k$-faces of $A$ at most the total volume of the $k$-faces of $B$?  What if, instead of just boxes, we allow $A$ and $B$ to be general convex polytopes?  (Or $A$ convex and $B$ non-convex, but I'm not really clear on how non-convex polytopes may be defined, so maybe let's ignore that.)
The case $k=n$ is trivial as that's just the volume.
The case $k=n-1$... well, I'm not very familiar with convex geometry, but I think there might be some general theorem that if a convex body $A$ is contained in a convex body $B$, then the surface volume of $A$ is at most the surface volume of $B$, right?  Or no?  I'm hoping that's true.  Well, assuming it's true, it'll handle $k=n-1$.  (If it is true, a reference would be appreciated!)
The link above shows how to handle $k=1$ in the box case (specifically $k=1$ and $n=3$, but the proof generalizes to higher $n$), and actually I think it might generalize to convex polytopes in general, but I haven't fully checked.  Unfortunately, as best I can tell, this proof doesn't seem to generalize to higher $k$.
The case $k=0$ is trivially true for the box case and trivially false for the more general case, which is why I've excluded it!
So what about other cases?  Like the smallest case not covered by the above (assuming the above are all correct) is $k=2$ and $n=4$.  Is it true for boxes?  For convex polyhedra more generally?  Does anybody know about this?
Thank you all!
 A: The statement is true for any $k$ when $A$ and $B$ are $n$-dimensional parallelograms.
To see this, we need a description of the volume of $k$-dimensional parallelograms in $\mathbb R^n$. Let $\{e_i\}$ be the standard basis for $\mathbb R^n$, so the wedge product $\wedge^k\mathbb R^n$ has a basis
$$
  \{e_{i_1}\wedge\ldots\wedge e_{i_k}\mid 1\leq i_1<i_2<\ldots<i_k\leq n\}.
$$
There is a unique norm on $\wedge^k\mathbb R^n$ such that the above basis is orthonormal. Given vectors $x_1,\ldots,x_k\in\mathbb R^n$, the volume of the parallelogram spanned by these vectors is
$$
  V(x_1,\ldots,x_k)=\|x_1\wedge\ldots\wedge x_k\|_2.
$$
(We can take this as a definition, but to see that it makes sense, check that it is invariant under a matrix rotating in the plane $\mathrm{span}\{e_1,e_2\}$, and that it reduces to the determinant when all the $x_i$ are in $\mathrm{span}\{e_1,\ldots,e_k\}$).
Now consider the $n$-dimensional parallelograms
$$
  A=\{a+t_1x_1+\ldots+t_nx_n\mid t_i\in[0,1]\},
$$
$$
  B=\{b+t_1y_1+\ldots+t_ny_n\mid t_i\in[0,1]\}.
$$
Let $X$ and $Y$ be the matrices whose columns are $x_i$ and $y_i$ respectively. We may suppose $Y$ is invertible (otherwise expand $B$ a bit and take the limit). Let $Z=Y^{-1}X$, so
$$
  x_i=\sum_j y_jz_{ji}.
$$
Fix $l$. Let $I=\{i\mid z_{li}>0\}$ and $I'=\{i\mid z_{li}\leq0\}$. Note that
$$
  a+\sum_{i\in I}x_i=a+\sum_jy_j\sum_{i\in I}z_{ji}\in B,
$$
$$
  a+\sum_{i\in I'}x_i=a+\sum_jy_j\sum_{i\in I'}z_{ji}\in B.
$$
Looking at the coefficient of $y_l$, we see that
$$
  1\geq\sum_{i\in I}z_{li}-\sum_{i\in I'}z_{li}=\sum_i|z_{li}|.
$$
This holds for each $l$. Now consider a $k$-dimensional face of $A$ spanned by $x_i$ for $i\in I$. The volume of this face is
$$\begin{eqnarray*}
  \|\wedge_{i\in I}x_i\|_2
    &=&\left\|\wedge_{i\in I}\left(\sum_j y_jz_{ji}\right)\right\|_2\\
    &=&\left\|\sum_J\wedge_{j\in J}y_j\det(Z_{JI})\right\|_2\\
    &\leq&\sum_J\|\wedge_{j\in J}y_j\|_2|\det(Z_{JI})|
\end{eqnarray*}$$
where the sum is over sets of indices $J$ of size $k$, and $Z_{JI}$ denotes the submatrix of $Z$ with rows from $J$ and columns from $I$. For fixed $J=\{j_1,\ldots,j_k\}$, we have
$$\begin{eqnarray*}
  \sum_I|\det(Z_{JI})|
    &\leq&\sum_{i_1,\ldots,i_k}|z_{j_1i_1}\ldots z_{j_ki_k}|\\
    &=&\prod_{j\in J}\sum_i|z_{ji}|\\
    &\leq&1.
\end{eqnarray*}$$
Therefore
$$
  \sum_I\|\wedge_{i\in I}x_i\|_2\leq\sum_J\|\wedge_{j\in J}y_j\|_2
$$
as required.

The statement is also true for $k=n-1$ when $A$ and $B$ are any convex polytopes; here is a sketch. Let $A_l$ denote the union of $l$-faces of $A$, and similarly for $B$. Given $x\in A_{n-1}\setminus A_{n-2}$, the ray from $x$ normal to the face containing $x$ intersects $B_{n-1}$ at a unique point $\phi(x)$. The resulting map
$$
  \phi:A_{n-1}\setminus A_{n-2}\to B_{n-1}
$$
is injective since $A$ is convex. Moreover if $\phi(x)\notin B_{n-2}$ then $\phi$ is linear and volume-increasing on some neighborhood of $x$. Since $A_{n-2}$ and $\phi^{-1}(B_{n-2})$ are of dimension $\leq n-2$, excluding these won't affect $(n-1)$-dimensional volume. Therefore
$$
  V(A_{n-1})=V(\phi^{-1}(B_{n-1}\setminus B_{n-2}))
    \leq V(B_{n-1}\setminus B_{n-2})=V(B_{n-1}).
$$
