Problem 1072

7. By symmetry, without loss of generality, assume $x \geqslant y \geqslant z \geqslant 0$. From $(x+y)+z=2 \times \frac{1}{2}$, we know that $x+y, \frac{1}{2}, z$ form an arithmetic sequence. Hence, let $x+y=\frac{1}{2}+d, z=\frac{1}{2}-d$. From $x+y \geqslant 2 z$, we get $\frac{1}{6} \leqslant d \leqslant \frac{1}{2}$, then $x y+y z+z x-2 x y z=(x+y) z+x y(1-2 z)=\frac{1}{4}-d^{2}+2 d x y \geqslant 0$. The equality holds if and only if $x=1, y=z=0$.
$$\begin{array}{l} \text { Also, } \frac{1}{4}-d^{2}+2 d x y \leqslant \frac{1}{4}-d^{2}+2 d\left(\frac{x+y}{2}\right)^{2}=\frac{1}{4}-d^{2}+\frac{1}{2} d\left(\frac{1}{2}+d\right)^{2}=\frac{1}{4}+\frac{1}{4} \times 2 d\left(\frac{1}{2}-d\right)^{2} \\ \leqslant \frac{1}{4}+\frac{1}{4}\left[\frac{2 d+\left(\frac{1}{2}-d\right)+\left(\frac{1}{2}-d\right)}{3}\right]^{3}=\frac{1}{4}+\frac{1}{4} \times \frac{1}{27}=\frac{7}{27} . \end{array}$$

The equality holds if and only if $x=y=z=\frac{1}{3}$. Therefore, $0 \leqslant x y+y z+z x-2 x y z \leqslant \frac{7}{27}$.