Find the maximum value of $u=x y z(x+y+z)$.

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When $x=y=z=\sqrt[3]{4}-1$, $u-3(\sqrt[3]{4}-1)^{4}$. The following proof shows: $u \leqslant 3(\sqrt[3]{4}-1)^{4}$. By the mean inequality,
$$
(x+y+z)^{4}=(x+y+z)^{3}(x+y+z) \geqslant 27 x y z(x+y+z)=27 u,
$$

Therefore,
$$
x+y+z \geqslant \sqrt[4]{27 u}.
$$

Also, by the semi-mean inequality,
$$
\begin{array}{l} 
x y z+\frac{(\sqrt[3]{4}-1)^{2}}{3}(x+y+z) \geqslant 2 \sqrt{\frac{(\sqrt[3]{4}-1)^{2}}{3} x y z(x+y+z)} \\
=2 \sqrt{\frac{(\sqrt[3]{4}-1)^{2}}{3}} u, \\
x y+y z+z x \geqslant \sqrt{3 x y z(x+y+z)}=\sqrt{3 u},
\end{array}
$$

By (1) $\times\left(1-\frac{(\sqrt[3]{4}-1)^{2}}{3}\right)+$ (2) + (3), and combining with the given conditions, we get
$$
3 \geqslant\left(1-\frac{(\sqrt[3]{4}-1)^{2}}{3}\right) \sqrt[1]{27 u}+2 \sqrt{\frac{(\sqrt[3]{4}-1)^{2}}{3} u}+\sqrt{3 u},
$$

Thus, $u \leqslant 3(\sqrt[3]{4}-1)^{4}$. Therefore, the maximum value of $u=x y z(x+y+z)$ is $3(\sqrt[3]{4}-1)^{4}$.