>>11455494x^2+y^2+z^2=a
(x-b)^2+y^2+x^2=a+b^2-2bx
if x>b, then x=b+c where c is positive
b^2-2*(b^2+bc) = -b^2-2bc
thus (b^2-2bx) is negative
therefore (x-b)^2+y^2+x^2=a-d where d is positive
thus (x-b)^2+y^2+x^2 < a
thus ((x-b)^2+y^2+x^2)^0.5 < a^0.5
therefore ANY subtraction from the largest extent of a cuboid will reduce the distance
therefore A->B is the biggest distance
