Given:
x,y,z > 0
xy+yz+zx <= 3/4
and
P=x+y+z+(1/x)+(1/y)+(1/z)
Find the Minimum Value of P.
Find x, y, and z when P = Minimum Value.
Haven't even finished the last one yet!
Anyway, the maximum value is either floating out in space, or it's on the boundary of this region. If it's a local minimum floating out in space, then grad[P(X, Y, Z)] = {0, 0, 0} for some X, Y, Z.
In general, grad[P(x, y, z)] = {1 - x^-2, 1 - y^-2 , 1 - z^-2}. A zero gradient only occurs at {1, 1, 1}, which is not within the region stated. Oh well, so much for the easy solution.
The solution is not at a point near the x = 0, y = 0, or z = 0 planes, because P would approach infinity as any of these variables got close to zero.
If it's on the surface xy+yz+zx = 3/4, then we can treat the problem as a constrained extremum. Let g(x,y,z) = xy + yz + zx - 3/4. The constrained extremum occurs when grad[P(x,y,z)] = Lgrad[g(x,y,z)] for some L.
grad[g(x,y,z)] = {y+z, x+z, x+y}.
This gives us four equations with four unknowns:
1-x^2 = L(y + z)
1-y^2 = L(x + z)
1-z^2 = L(x + y)
xy + yz + zx = 3/4.
Fortunately, I already had my suspicious about this problem, and I suspected that the answer was {1/2, 1/2, 1/2}. This turns out to be part of the solution for the system of equations, with L=3/2. I haven't proved that this is a minimum, but inspection shows us that it's the only critical point on the surface for positive x, y, and z. Moreover, if x, y, or z moves toward zero or infinity, P goes to infinity. So I'm confident this point is a minimum.
My answer is P = 7 1/2, at {1/2, 1/2, 1/2}.
Evil!
