Lagrange Multipliers - Solving system of equation $$\begin{array}{ll} \text{extremize} & \overbrace{xy + xz + yz}^{=: f(x,y,z)}\\ \text{subject to} & x^2 + y^2 + z^2 = 1\end{array}$$

I get the following Lagrange system of equations
$y + z - 2x\lambda = 0\qquad(1)$
$x + z - 2y\lambda = 0\qquad(2)$
$x + y - 2z\lambda = 0\qquad(3)$
$x^2 + y^2 + z^2 = 1\qquad(4)$
if i do (1) + (2) + (3) and divide by 2 i get $x +y+z = \lambda(x + y + z)$ which means $\lambda = 1$ or $x+y+z=0$
I can properly conclude the $\lambda = 1$ case but i'm not being able to develop the case $x+y+z=0$. Any tips on how to procede?

$$1-(xy+xz+yz)=\sum_{cyc}(x^2-xy)=\frac{1}{2}\sum_{cyc}(x-y)^2\geq0,$$
which says that $$xy+xz+yz\leq1.$$
The equality occurs for $x=y=z=\frac{1}{\sqrt3},$ which says that we got a maximal value.
Also, (it's  the case, for which you look) $$\frac{1}{2}+xy+xz+yz=\frac{1}{2}\sum_{cyc}(x^2+2xy)=\frac{1}{2}(x+y+z)^2\geq0,$$
which says $$xy+xz+yz\geq-\frac{1}{2}.$$
The equality occurs for $z=0$, $x=\frac{1}{\sqrt2}$ and $y=-\frac{1}{\sqrt2},$ which says that we got a minimal value.
