Given the function $$f(x,y,z)=xy+xz+yz$$ I have to find its extremes on the set $E=\{(x,y,z)\in\mathbb R|x^2+y^2+z^2=1\}$.
Given the fact that $E$ is the border of a sphere, I can conclude that it is a compact set and because $f$ is continuous then there will be some extremes.
I proceed with the system of equations given by $\nabla f=\lambda\nabla g$ and the constraint $g(x,y,z)=x^2+y^2+z^2-1$ obtaining
$$\cases{y+z=2\lambda x\\x+z=2\lambda y\\x+y=2\lambda z\\x^2+y^2+z^2=1}\Rightarrow\cases{x+y+z=(2\lambda+1) x\\x+y+z=(2\lambda+1) y\\x+y+z=(2\lambda+1) z\\x^2+y^2+z^2=1}$$
Then for $\lambda\ne-\frac 1 2$ I obtain $x=y=z$ and putting it in the last equation I obtain that the points of extreme are in $x=y=z=\pm\frac 1 {\sqrt 3}$ and then I can analyze these points.
If on the other hand $\lambda=-\frac 1 2$ using the first system of equations I obtain
$$\cases{y+z=-x\\x+z=-y\\x+y=-z\\x^2+y^2+z^2=1}\Rightarrow\cases{x+y+z=0\\x^2+y^2+z^2=1}$$ which leads to $x=-y-z$ and produces a conic with equation $y^2+z^2+yz=\frac 1 2$. I think that means that the points of extreme of $f$ in this case are on that conic, but then how can I analyze these points?
On the intersection of the plane $x+y+z=0$ and the unit sphere, $$xy+xz+yz=\frac12\left((x+y+z)^2-(x^2+y^2+z^2)\right)=\frac{0-1}2=-\frac12.$$