since xz+yx+yz = 3
x^2 +y^2 +z^2 +2(xz+yx+yz) >6 on the surface
Consider the problem of finding the points on the surface that are closest to the origin.
1) Use the identity to prove that is not equal to 0 for any point on the given surface.
2) Use the method of Lagrange multipliers to find a system of four equations in and whose solutions will give the closest points.
3) Find the points on that are closest to the origin.
I'm clueless on what to do for the 1st part (although I imagine it's actually something simple), but I think I have the second part down. Problem is, I think I probably need to use the 1st part for the 3rd somehow.
For the second part, I found the system of four equations to be:
you should be able to show
x^2+ xz = y^2 + yz
(x-y)(x+y) = -z(x-y)
if x doesn't equal y then x+y = -z but this yields x+y+z = 0
but this can't happen so x = y
similarly you can show y^2 +xy = z^2 + xz
and use the same argument so x = y = z
use this in your constraint xy + xz +yz =3 to solve