Assume that I have a parallelepiped $A$ that has the following dimensions: $w_{A}, l_{A}, h_{A}$ s.t. $w_{A}+l_{A}+h_{A}= 115$. Consider another parallelepiped $B$ with dimensions $w_{B}, l_{B}, h_{B}$ such that $w_{B}+l_{B}+h_{B}>115$.
I want to prove I cannot fit $B$ inside $A$ using the notion of width $\sup\limits_{x,y\in A}\langle s, x-y\rangle$ with $s \in \mathbb S^{2}$.
My attempt:
define $w(A)= \sup\limits_{x,y\in A}\langle s_{w(A)}, x-y\rangle $ with $s_{w(A)}\in \mathbb S ^{2}$.
Assuming that $B$ fits into $A$, then it must follow that $w(B)=\sup\limits_{x,y\in B}\langle s_{w(A)}, x-y\rangle\leq w(A)$.
Could I not just continue in the way as above to show that $l(B) \leq l(A)$ and $h(B) \leq h(A)$ but this seems too straightforward and that I am not using anything actually.
I additionally thought of using the following properties of the width, height and length as defined above to get some contradiction:
$w(A+B)= w(A)+w(B)$ and $w(\lambda A)= \lvert\lambda\rvert w(A)$
Any ideas on how to go about proving the above?