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A different solution to the Box in box problem, with less math:

Consider the square of (length + with + height). The sum of the diagonal terms is the square of the distance between opposite corners, the sum of the off diagonal terms equals half the area of the boxes. We can show that both these terms are smaller for the innermost box.

It’s obvious for the diagonal terms, as the opposite corners of a box are the two points of the box that are furthest apart. So the opposite corners of the inner box must be separated by less than the opposite corners of the outer one.

For the cross diagonal terms: take axes that are aligned with the inner box. Consider the parts of the outer box that would be projected on the inner box if you projected either on the x&y, x&z, or y&z planes (i.e. you project on the faces of the inner box). These six pieces of the outer box don’t intersect, and each of them is larger than the face of the inner box on which they project. So this part of the outer box has an area bigger than the total are of the inner box. QED