First off clearly (0,0,1), (0,1,0), and (1,0,0) yield 0,
(1/3,1/3,1/3) yields 7/27. Not that this proves anything.
y = 1-x-z so we can substitute to give the expression
x(1-x-z) + xz + (1-x-z)z - 2x(1-x-z)z
this can be rewritten as a quadratic in x:
(2z-1)x^2 + (2z^2-3z+1)x + (z-z^2)
(2z-1)x^2 + (2z-1)(z-1)x + (z-z^2)
Which has a min/max at -b/2a = -(z-1)/2
If z=1/2 this expression just evaluates to 1/4 (clearly not a min or max)
If z>1/2 this is always an upward parabola in x (case 1)
If z<1/2 this is always a downward parabola in x (case 2)
Substituting -(z-1)/2 for x gives, with some simplifying
-3z^2/2+z/2 is zero when z=0 or z=1/3
So by case 2 this is a maximum value, case 1 does not apply.
Putting 1/3 into the cubic gives 7/27.
Posted by Jer
on 2013-03-20 12:54:06