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?



Two rectangular parallelepiped
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.


