$9 \cdot 59$ Let $x_{i} \geqslant 0, i=1,2, \cdots, n, n \geqslant 2$ and $\sum_{i=1}^{n} x_{i}=1$, find the maximum value of $\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right)$.
[Solution] Let $z=\left\{\left(x_{1}, x_{2}, \cdots, x_{n}\right) ; x_{i} \geqslant 0, i=1,2, \cdots, n\right.$ and $\sum_{i=1}^{n} x_{i}=$ $1\}, F(v)=\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right)$. When $n=2$, it is obvious that
$$\max _{v \in:} F(v)=\frac{1}{4}$$

and it is achieved when $x_{1}=x_{2}=\frac{1}{2}$. For the case $n \geqslant 3$, take any $v=\left(x_{1}, x_{2}, \cdots, x_{n}\right) \in z$, without loss of generality, assume $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}$. Let
$$w=\left(x_{1}, x_{2}, \cdots, x_{n-1}+x_{n}, 0\right),$$

then $w \in z$ and
$$\begin{aligned}
F(w)= & \sum_{1 \leqslant i<j \leqslant n-2} x_{i} x_{j}\left(x_{i}+x_{j}\right)+\sum_{i=1}^{n-2} x_{i}\left(x_{n-1}+x_{n}\right)\left(x_{i}+x_{n-1}+x_{n}\right) \\
= & \sum_{1 \leqslant i<j \leqslant n-2} x_{i} x_{j}\left(x_{i}+x_{j}\right)+\sum_{i=1}^{n-2} x_{i} x_{n-1}\left(x_{i}+x_{n-1}\right) \\
& +\sum_{i=1}^{n-2} x_{i} x_{n}\left(x_{i}+x_{n}\right)+2 x_{n-1} x_{n} \sum_{i=1}^{n-2} x_{i} .
\end{aligned}$$

Thus,
$$F(w)=F(v)+x_{n-1} x_{n}\left(2 \sum_{i=1}^{n-2} x_{i}-x_{n-1}-x_{n}\right)$$

Since $n \geqslant 3, x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}$, we have
$$\sum_{i=1}^{n-2} x_{i} \geqslant \frac{1}{3}, \quad x_{n-1}+x_{n} \leqslant \frac{2}{3}$$

Therefore, $F(w) \geqslant F(v)$. By induction, it is easy to see that for any $v \in z$, there exists $u=(a, b, 0, \cdots, 0) \in z$ such that
$$F(u) \geqslant F(v)$$

Thus, the maximum value sought is $\frac{1}{4}$, and it is achieved when $x_{1}=x_{2}=\frac{1}{2}, x_{3}=\cdots=x_{n}=0$.