Proposed by me We have $-1 = -(x^2+y^2+z^2) \leq xy + yz - xz \leq \frac{x^2+y^2+z^2}{2} = \frac{1}{2}$, as the left one is as $(x+y)^2 + (y+z)^2 + (x-z)^2 \geq 0$ while the right one is $(x+z-y)^2 \geq 0$. In the first case equality holds only of $x=z = -y = \frac{1}{\sqrt{3}}$, while in the second one it holds e.g. for $x=0, y=z= \frac{1}{\sqrt{2}}$. To show equality cannot happen over the rationals, note that we must have $y=x+z$ and in $x^2 + z^2 + (x+z)^2 = 1$ writing $x = \frac{p}{r}$, $z = \frac{q}{r}$ yields $p^2 + q^2 + (p+q)^2 = r^2$, which has no non-zero solution by mod $3$ (or mod $4$) combined with infinite descent.