[jmblpati]

I think there's an easier to visualize solution to the box problem.

Let A have dimensions (a,b,c) and let B have dimensions (x,y,z). Assume A fits inside B.

We have (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc. This is the sum of the A's hypotenuse squared plus its surface area. The same holds for B.

Note that A's hypotenuse is at most that of B-- the hypotenuse of a is its longest axis, and it needs to fit in B somehow. Further, note that the surface area of A is less than that of B. To see this, consider the nesting of A inside B and realize that both boxes' interiors are convex sets. Imagine inflating A inside of B by taking the sets A_t consisting of all points within B that are within distance t of a point in A. It is not hard to see that this inflating operation can only increase the surface area of A, and since the maximum surface area we can get is that of B we have that A has smaller surface area than that of B. Thus,

(a+b+c)^2 = (A hypotenuse)^2 + (A surface area) <= (B hypotenuse)^2 + (B surface area) = (x+y+z)^2. The claim follows.