Given $x, y, z \geq 0, x+y+z=4$, find the maximum value of $x^{3} y+y^{3} z+z^{3} x$.

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One, Solution: By the cyclic property, we may assume $x=\max \{x, y, z\}$.
(1) If $x \geq y \geq z$, then
$$
x^{3} y+y^{3} z+z^{3} x \leq(x+z)^{3} \cdot y \leq \frac{1}{3}\left(\frac{3(x+z)+3 y}{4}\right)^{4}=27
$$

Equality holds when $x=3, y=1, z=0$;
(2) If $x \geq z \geq y$, similarly we have
$$
x^{3} z+y^{3} x+z^{3} y \leq(x+y)^{3} \cdot z \leq \frac{1}{3}\left(\frac{3(x+y)+3 z}{4}\right)^{4}=27
$$

And $x^{3} y+y^{3} z+z^{3} x \leq x^{3} z+y^{3} x+z^{3} y \Leftrightarrow x^{3}(z-y)+y^{3}(x-z) \geq z^{3}(x-z)+z^{3}(z-y)$, which holds by $x \geq z \geq y$, hence $x^{3} y+y^{3} z+z^{3} x \leq x^{3} z+y^{3} x+z^{3} y \leq 27$
Combining (1) and (2), the maximum value of the original expression is 27, equality holds if and only if $x=4, y=1, z=0$ and its cyclic permutations.