Prove that if we have two rectangular parallelepiped (cuboids) such that one of them is placed inside the other then the sum of the three lengths of the inner parallelepiped is at most the sum of the three lengths of the exterior parallelepiped. In 2 dimensions the problem is trivial. Does this hold in higher dimensions? There is a way to prove it in higher Euclidean dimensions?
This version works for all parallelepipeds, not only rectangular ones:
If you replace each parallelepiped by all points that have distance at most $\varepsilon$ to a point in the parallelepiped, you can still place the smaller inside the bigger one. In particular, the smaller object has a smaller volume.
We divide up the extended parallepipeds by extending the planes corresponding to the six faces. This gives the volume of the original solid in the center, parallelepipeds of height $\varepsilon$ on top of each face, partial cylinders (with slanted parallel ends) of radius $\varepsilon$ and length the corresponding edge, and partial spheres of radius $\varepsilon$ around each vertex.
The partial spheres add up to exactly one whole sphere simply by translation. The partial cylinders corresponding to parallel edges add up to one whole cylinder by translation.
Now let $\varepsilon$ tend to infinity (yes, really). The term with $\varepsilon^3$ comes from the sphere of radius $\varepsilon$ and does not depend on the parallelepiped at all, . The coefficient of $\varepsilon^2$ comes from the cylinders and is clearly the sum of the edges times some constant.
So, for the inequality to be valid the sum of the edges of the smaller box must be smaller than the sum of the edges of the larger box.
I don't see any problem with the generalization to higher dimension.

$\begingroup$ I think this works, at least in three dimensions, but with some corrections: the term with $\epsilon^3$ is just the volume of a sphere of radius $\epsilon$, independent of the number of vertices: the spherical caps on each vertex form a dissection of a whole sphere. The $\epsilon^2$ term for an edge depends on its dihedral angle, but the twelve edges can be grouped into sets of four such that within each set the lengths are equal and the complementary dihedrals add to 2π, so for the parallelepiped (but not for other polyhedra) the total depends only on length. $\endgroup$ – David Eppstein Oct 31 '11 at 21:48

$\begingroup$ @David Eppstein: What you write is absolutely correct, but note that in the OP we had rectangular parallelepipeds, so I mentioned the number of vertices to indicate that it is not necessary to think more about this in this case. Sorry for being both too lazy to repeat the word "rectangular" and to give the more general argument. I don't like the English term "rectangular parallelepiped". $\endgroup$ – user11235 Oct 31 '11 at 21:56

3$\begingroup$ This is an old chestnut. I first heard this problem from Isaac Kornfeld, told it to Bela Bollobas who included in his book of coffee time problems. Since airlines (and the post office) charge for packages on the basis of the
linear dimension' (i.e. the sum of the dimensions), this question can also be phrased
can you cheat and get a cheaper price by enclosing the smaller box in a larger one?'. $\endgroup$ – Anthony Quas Oct 31 '11 at 22:27 
1$\begingroup$ This is a really nice argument. It does work for all parallelepipeds and in all dimensions. To see that everything fits perfectly around edges and virtices it helps to look at $\epsilon$neighborhoods of virtices and lines in $\mathbb R^n$ with the $\mathbb Z^n$ action generated by the translations along edges of the parallellepiped. $\endgroup$ – Vitali Kapovitch Nov 1 '11 at 1:54

$\begingroup$ @Vitali: Indeed this is a very nice proof. Thanks! $\endgroup$ – heartwork Nov 2 '11 at 13:01
This is true for 3dimensional parallelepipeds. Let the sides of the ambient parallelepiped be $A,B,C$ and of the inner one $a,b,c$. Then it's clear that $A^2+B^2+C^2\ge a^2+b^2+c^2$ since the diameter of the ambient body is obviously bigger. Also, it's easy to see that the there is a 1Lipschitz map of $\mathbb R^3$ onto the smaller parallelepiped (just take the nearest point projection map). Being 1Lipschitz this map does not increase 2dimensional area and therefore, applying this to the surface of the bigger parallellepiped we get $2(AB+BC+AC)\ge 2(ab+bc+ac)$. Adding the above two inequalities we get $(A+B+C)^2\ge (a+b+c)^2$.
I learned of this problem from Anton's Petrunin's list of exercises in orthodox geometry (which has a lot of other cool problems btw).
I don't know if the same holds in higher dimensions.

$\begingroup$ Nice. I don't see what part fails for dimension $4$ and higher. $\endgroup$ – Noam D. Elkies Oct 31 '11 at 16:06

1$\begingroup$ @Noam: you need to prove only that the projection of 2skeleton is ONTO. But it seems to be true. $\endgroup$ – Ilya Bogdanov Oct 31 '11 at 16:17

1$\begingroup$ @Noam D. Elkies. This argument doesn't work already in dimension 4 (at least I don't see how to make it work) because we need an inequality on 2dimensional area of the 2skeletons and the projection map doesn't preserve them. But the argument by thei below works in all dimensions and for all parallelepipeds, not just for rectangular ones. $\endgroup$ – Vitali Kapovitch Nov 1 '11 at 1:59
What if we ask the same problem with tetrahedron instead of parallelepiped. Consider two tetrahedrons one placed inside the other. Prove that the sum of all 6 sides of inner tetrahedron is at most the sum of the 6 sides of exterior tetrahedron. This looks much harder!

$\begingroup$ Please don't add comments as answers. If you want to ask a different question, ask it as a different question. $\endgroup$ – Yemon Choi Dec 17 '11 at 8:53

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