The length of a line segment can't be greater than the sum of its projections onto all three axes. Therefore the "perimeter" of the inner box can't be greater than the sum of its projections. But that sum is bounded from above by the "perimeter" of the outer box (which is equal to the sum of its own projections, because the outer box is axis-aligned).
I'm having trouble with your "therefore" conclusion. The projection of the perimeter of the inner box could easily be shorter than the sum of the lengths of the projections of each edge. It seems like you need a less obvious (to me, anyway) statement about the projection of a collection of lines.
By projected perimeter I mean the sum of lengths of projections of all edges. I'm not canceling them out or anything. Think of the box as a graph, its projection is another graph that happens to lie on a straight line, but we can measure the sum of its edges regardless.
It's not completely obvious why the projected perimeter of the inner box is bounded by the projected perimeter of the outer box, but it's a statement about one-dimensional segments that's easy enough to check.
I'm still not quite following. Is the "perimeter" here the sum of the lengths of projections of the edges, then how is that different from "the sum of its projections"?
Anyway if you define "perimeter" that way then for the outer box the "perimeter" is in fact 4 * (a+b+c), where a, b, c are the dimensions of the box. I also agree that for the inner box the "perimeter" is at least 4 * (a'+b'+c'), by the triangle inequality.
What is not at all clear to me is why the "perimeter" of the inner box is no larger than the "perimeter" of the outer box in this setup. You say it's easy to check, but it doesn't seem very obvious to me.
Let's say the projections of the inner box's edges onto the X axis have three distinct lengths a,b,c (all positive). Also let's say the distance between the leftmost and rightmost projected vertices is d. The nontrivial fact is that d=a+b+c, and not say a-b+c. That leads to the desired inequality, I think.
Ah, I see. Yes, I agree that given that fact you get the desired inequality.
But this fact is, as you say, nontrivial. It's certainly false for various non-box-like shapes afaict, so there's something special to boxes that needs proving here.
Just note that all combinations ±a±b±c can be realized by paths on the box.
I'm sorry, I see now that my original comment glossed over tons of stuff that was clear only to me. Instead of saying "easy to check" I should've posted my napkin sketch with the easy check.
Anyway, my solution is still simpler and more natural than the one in that pdf :-)
