Question 389078
Without loss of generality, let the diameter of the sphere be {{{sqrt(3)}}} (you'll see why I picked such a strange number), and that the dimensions of the rectangular solid are x, y, z. Given that:


{{{x^2 + y^2 + z^2 = (sqrt(3))^2 = 3}}}


we want to show that for all positive x, y, z that


{{{xyz <= 1}}} (where 1 is the equality case and the volume of an inscribed cube).


Using the AM-GM inequality,


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


Since {{{x^2 + y^2 + z^2 = 3}}}, this becomes


{{{1 >= root(3, x^2y^2z^2)}}}


Cubing both sides,


{{{1 >= x^2y^2z^2}}}


{{{1 >= xyz}}}


{{{xyz <= 1}}}, as desired. Note that the equality case is where x = y = z = 1, which is also the equality case of AM-GM.


There is probably an optimization solution involving derivatives, however, it would involve several variables and would be fairly tedious. I'm sure many other solutions exist. This solution is the first one that came to my mind.