10 Let $x, y, z \geqslant 0$, and $x+y+z=1$, find the maximum and minimum values of $x^{2} y+y^{2} z+z^{2} x$.


10. First, $x^{2} y+y^{2} z+z^{2} x \geqslant 0$, when $x=1, y=z=0$ the equality holds, so the minimum value is 0. Without loss of generality, assume $x=\max \{x, y, z\}$, then
(1) When $x \geqslant y \geqslant z$, we have $x^{2} y+y^{2} z+z^{2} x \leqslant x^{2} y+y^{2} z+z^{2} x+z[x y+$ $(x-y)(y-z)]=(x+z)^{2} y=(1-y)^{2} y \leqslant \frac{4}{27}$.
(2) When $x \geqslant z \geqslant y$, by $(x-y)(y-z)(z-x)=\left(x y^{2}+y z^{2}+z x^{2}\right)-$ $\left(x^{2} y+y^{2} z+z^{2} x\right)$ we know $x^{2} y+y^{2} z+z^{2} x \leqslant x^{2} y+y^{2} z+z^{2} x+(x-y)(y-z)$ $(z-x)=x^{2} z+z^{2} y+y^{2} x \leqslant \frac{4}{27}$. Therefore, the maximum value is $\frac{4}{27}$, and it is achieved when $x=\frac{2}{3}, y=\frac{1}{3}$, $z=0$.