A rectangular box $R$ with dimensions $A$, $B$, and $C$ contains another rectangular box $r$ with dimensions $a$, $b$, and $c$. Both boxes are rigid. Show that $A+B+C\geq a+b+c$. This implies that there is no way of getting cheaper postage by wrapping with another box when postage depends only on the sum of dimensions.
Proof:
It follows directly from two facts.
The first is $AB+AC+BC\geq ab+ac+bc$, or $R$'s surface area is no less than $r$'s. Project each face of $r$ outward onto $R$'s surface. The projections of $6$ faces on $R$ are disjoint and cannot be smaller than the original area.
The second is $A^2+B^2+C^2\geq a^2+b^2+c^2$. Imagine $r$ contains a rigid stick at its diagonal, i.e. with length $\sqrt{a^2+b^2+c^2}$. The stick has to fit in $R$ as well, which can take a stick no longer than $\sqrt{A^2+B^2+C^2}$.
Q.E.D.
What happens in higher dimensional space?
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