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 = 1xz so we can substitute to give the expression
x(1xz) + xz + (1xz)z  2x(1xz)z
this can be rewritten as a quadratic in x:
(2z1)x^2 + (2z^23z+1)x + (zz^2)
(2z1)x^2 + (2z1)(z1)x + (zz^2)
Which has a min/max at b/2a = (z1)/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 (z1)/2 for x gives, with some simplifying
z^3/2+z^2/4+1/4
It derivative
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 20130320 12:54:06 