Let $S⊂\mathbb R^3$, and $S_1,S_2,S_3⊂\mathbb R^2$ the projections planes $xy$, $yz$ and $xz$ of $S$ it is known that
$$V(S)^2≤A(S_1)A(S_2)A(S_3)$$

This also follows from the first inequality in this text
http://www.ma.huji.ac.il/~ehudf/docs/KKLBKKKL.pdf using the indicators functions of $S_1$,$S_2$ and $S_3$.

Now using this inequality we will prove that the cube has the least surface area.
Given $S\subset\mathbb R^3$ we can construct a box $B$ with projection $B_1,B_2,B_3$ such that $A(S_i)=A(B_i)$ just by taking the box with sides $$\sqrt{ A(S_i)A(S_j)/A(S_k)}$$ for  $i,j,k=1,2,3$ all different. Since for the box the inequality becomes equality it follows that $V(S)\leq V(B)$. Now take an orthogonal polyhedron $P$ and let $A(P)$ its surface area, then crearly $$A(P)\geq 2(A(P_1)+A(P_2)+A(P_3)),$$ then if we take the box with $A(B_i)=A(P_i)$ it follows that $A(B)=2(A(P_1)+A(P_2)+A(P_3))\leq A(P)$ therefore
$B$ increases the volume and decreases the surface area.
Finally it is an easy calculus exercise that from all the boxes with fixed area, the one with less surface area is the cube.