Question 389682
Without loss of generality, let the fixed volume be 1, and the dimensions of the rectangular solid be x, y, z. Then,


{{{xyz}}} = 1, and we want to show whether


{{{2(xy + yz + zx) >= 6}}} --> {{{xy + yz + zx >= 3}}} is true (if it is true, then the solid must have a surface area of at least 6, and the solid of least surface area is a cube).


Applying the AM-GM inequality,


{{{(xy + yz + zx)/3 >= root(3, x^2y^2z^2)}}}



Since {{{xyz = 1}}}, replace it into the right-hand side and simplify:


{{{(xy + yz + zx)/3 >= root(3, 1) = 1}}}


{{{xy + yz + zx >= 3}}} as desired. The equality of AM-GM occurs when x = y = z, i.e. the rectangular solid is a cube.