4.6.15 ** Let $x, y, z \geqslant 0$, and $x+y+z=1$. Find the minimum and maximum values of $x y+y z+z x-3 x y z$.


Assume $x \geqslant y \geqslant z$, then $0 \leqslant z \leqslant \frac{1}{3}, 1-3 z \geqslant 0$. Therefore,
$$
x y+y z+z x-3 x y z=x y(1-3 z)+y z+z x \geqslant 0,
$$

when $x=1, y=z=0$, the equality holds. Hence the minimum value is 0.
Now let $z=\frac{1}{3}-k$, then $0 \leqslant k \leqslant \frac{1}{3}, x+y=\frac{2}{3}+k$. Thus,
$$
\begin{aligned}
x y+y z+z x-3 x y z & =x y(1-3 z)+z(x+y) \\
& =x y(3 k)+\left(\frac{1}{3}-k\right)\left(\frac{2}{3}+k\right) \\
& \leqslant\left(\frac{x+y}{2}\right)^{2}(3 k)+\left(\frac{1}{3}-k\right)\left(\frac{2}{3}+k\right) \\
& =\left(\frac{1}{3}+\frac{k}{2}\right)^{2}(3 k)+\left(\frac{1}{3}-k\right)\left(\frac{2}{3}+k\right) \\
& =\frac{k}{3}+k^{2}+\frac{3}{4} k^{3}+\frac{2}{9}-\frac{k}{3}-k^{2} \\
& =\frac{3}{4} k^{3}+\frac{2}{9} \leqslant \frac{3}{4}\left(\frac{1}{3}\right)^{3}+\frac{2}{9}=\frac{1}{4},
\end{aligned}
$$

when $x=y=\frac{1}{2}, z=0$, the equality holds, hence the maximum value is $\frac{1}{4}$.