Proof: By symmetry, we may assume $x \geqslant y \geqslant z$, thus $1=x+y+z \geqslant 3 z$, which means $z \leqslant \frac{1}{3}$. Therefore, $2 x y z \leqslant \frac{2}{3} x y \leqslant x y$, hence
$$y z+z x+x y-2 x y z \geqslant 0$$

For the right inequality, we can consider using the mean substitution: Let $x+y=\frac{1}{2}+t, z=\frac{1}{2}-t$, then by $x+y \geqslant 2 z$ we get $\frac{1}{6} \leqslant t \leqslant \frac{1}{2}$, thus
$$\begin{array}{l} y z+z x+x y-2 x y z=\frac{1}{4}-t^{2}+x y-2 t \\ \leqslant \frac{1}{4}-t^{2}+2 t\left(\frac{x+y}{2}\right)^{2}=\frac{1}{4}-t^{2}+\frac{t}{2}\left(\frac{1}{2}+t\right)^{2} \\ =\frac{1}{4}+\frac{1}{4} \cdot 2 t \cdot\left(\frac{1}{2}-t\right)\left(\frac{1}{2}-t\right) \\ \leqslant \frac{1}{4}+\frac{1}{4}\left[\frac{2 t+\left(\frac{1}{2}-t\right)+\left(\frac{1}{2}-t\right)}{3}\right]^{3}=\frac{7}{27} . \end{array}$$