Question

Find the range of yz+zx+xz if x^2 + y^2 +z^2 =1

Answer 1

I didn't go through anything rigorously, but I'm thinking yz+zx+xz can range from -1 to 1.
Your restriction says that any point (x,y,z) is on the surface of a sphere of radius 1. So maximizing for any pair of (x,y,z) you'll get 1/3. Minimizing you'll get -1/3.
That's my thought.

Answer 2

we know that (x+y+z)^2 = x^2+y^2+z^2 +2(xy+yz+zx) >= 0
since x^2+y^2+z^2 = 1
thus,1+2(xy+yz+zx) >= 0
i.e. xy+yz+zx >= -1/2 ---------------------------(A)
Now we know that AM>= GM [Progression]
x^2+y^2+z^2 >= 3 (x^2y^2z^2)^1/3 ------(1)
similarly xy+yz+zx >= 3 (x^2y^2z^2)^1/3 ------(2)
thus from (1) and (2)
x^2+y^2+z^2 >=xy+yz+zx
clearly, xy+yz+zx <= 1 --------------------------------(B)

Therefore from (A) and (B) clearly,
Interval will be [-1/2.1]

Answer 3

Nice proof. However, you asked a different question originally - probably a typo. But for that typo-ed question, I'll stick to my original response.