Question 389523
This is actually somewhat similar to a previous solution I posted, in which the question was asking to prove that the rectangular solid of maximum volume inscribed in a sphere was a cube.


Let x, y, z be the dimensions of the rectangular solid, and without loss of generality let the diameter of the sphere be {{{sqrt(3)}}} (I assigned this number in the previous solution as well). From this, we establish that:


{{{x^2 + y^2 + z^2 = (sqrt(3))^2 = 3}}} (this follows from Pythagorean theorem)
Surface area = {{{2(xy + yz + zx)}}}
Such an inscribed cube has side length 1, and has surface area of 6.



We want to prove that


{{{2(xy + yz + xz) <= 6}}} --> {{{xy + yz + xz <= 3}}}


To show this, I used the Cauchy-Schwarz inequality (see below) letting a_1 = x, a_2 = y, a_3 = z, b_1 = y, b_2 = z, b_3 = x.


{{{(x^2 + y^2 + z^2)(y^2 + z^2 + x^2) >= (xy + yz + zx)^2}}}


Since {{{x^2 + y^2 + z^2 = 3}}}, the left hand side equals 9. Therefore we can substitute 9 and take the square root of both sides to obtain


{{{9 >= (xy + yz + zx)^2}}}


{{{3 >= xy + yz + zx}}} 


as desired. Note that the equality case occurs when x = y = z, i.e. the rectangular solid is a cube.


This is probably the easiest solution, there might be other solutions using optimization given a derivative of a function of two variables x,y (since z is determined from x,y). However it would be rather lengthy compared to this solution (it's pretty amazing that the Cauchy-Schwarz inequality produces the result immediately).



Note: the Cauchy-Schwarz inequality says that, for positive real numbers a_1, a_2, ..., a_n and b_1, b_2, ...b_n, then


{{{sum((a_i)^2, i = 1, n)*sum((b_i)^2, i = 1, n) >= (sum((a_i)(b_i), i = 1, n))^2}}} The equality case occurs when all a_i are equal to c*(b_i) where c is a constant.