Geometric proof that twice the square of the diagonal of a rectangular cuboid equals the surface area if and only if it is a cube

Question

In the answers to this question it is shown that the surface area of a cube equals twice the square of the length of its diagonal.

This is very much related to the following algebraic claim:

Three numbers $a,b,c\in \mathbb{R}$ satisfy $a=b=c$ if and only if $a^2 + b^2 + c^2 = ab + bc + ac$.

Given a rectangular cuboid with sides $a,b,c$, the LHS of the equality is the square of the length of the diagonal. The RHS of the equation is half of the surface area. Therefore, A geometric formulation of the algebraic claim would be:

The surface area of a rectangular cuboid equals twice the square of the length of its diagonal if and only if it's a cube.

I would like to know if anyone can come up with a geometric proof of this claim. To start with, I don't even have any geometric intuition for why this equality is correct for a cube.

Answer 1

For a rectangle with diagonal $d$ you have (see figure): $$ 2\cdot\text{area of rectangle}=\text{area of rhombus}\le d^2, $$ where equality holds only if the rhombus is a square, i.e. if the rectangle is a square.

Let now $A_1$, $A_2$, $A_3$ be the areas of three concurring faces of a cuboid, with diagonals $d_1$, $d_2$, $d_3$. By the above remark we have: $$ \text{surface area of cuboid}=2A_1+2A_2+2A_3\le d_1^2+d_2^2+d_3^2=2(\text{diagonal of cuboid})^2, $$ where equality holds only if all faces are squares.

Answer 2

In cuboid $AG$ join $AG$ and $AC$. By Pythagorean theorem, $AG^2=CG^2+AC^2$, and $AC^2=CD^2+AD^2=CD^2+CB^2$. Therefore$$AG^2=CG^2+CD^2+CB^2$$and$$2AG^2=2CG^2+2CD^2+2CB^2$$But the surface area of the cuboid is$$2CG\times CD+2CG\times CB+2CD\times CB$$and this equals $2AG^2$ only if $CG=CD=CB$, i.e. only if the faces are squares and the cuboid is a cube.