Suppose you want to send a package (a box (an orthohedron)). The delivery company tells you the price of the package will be the sum of the dimensions of the box, i.e., if it has sides $a \times b \times c$ then it will cost $a+b+c$ $ to send. Show you can't fit a more expensive box into a cheaper one.
NOTE: I have tried a few different approaches and I think I have a proof for the planar case, though rather complicated, but still should be adaptable to the 3D case. What I'm looking for (mostly) are elegant solutions or ideas that don't involve a lot of computations / analysis tricks.
NOTE 2: It's easy to show, however, that you can achieve a better price for your mailing with several packages. This is not the case here, we just consider exactly one inner box.
This is an old problem. Let $B$ be a box, and define, for $r>0$, the $r$-neighbourhood $B^{(r)}$ to be the set of points of Euclidean distance at most $r$ from $B$. If $B_1\subseteq B_2$ then $B_1^{(r)}\subseteq B_2^{(r)}$. Now consider the volumes of the $B_i^{(r)}$ as $r\to\infty$.