$9 \cdot 67$ Let $n$ be a fixed integer, $n \geqslant 2$.
(a) Determine the smallest constant $c$, such that the inequality
$$\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) \leqslant c\left(\sum_{1 \leqslant i \leqslant n} x_{i}\right)^{4}$$

holds for all non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n} \geqslant 0$.
(b) For this constant $c$, determine the necessary and sufficient conditions for equality to hold.
[Solution] Since the inequality is homogeneous and symmetric, we may assume without loss of generality that $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n} \geqslant 0$, and
$$\sum_{i=1}^{n} x_{i}=1$$

Thus, we only need to consider
$$F\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\sum_{1 \leqslant i < j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right)$$

Therefore, when adjusting $x$ to $x^{\prime}$, the function value $F$ strictly increases. For any $x=\left(x_{1}, \cdots, x_{n}\right)$, after several adjustments, we eventually get
$$\begin{aligned}
F(x) & \leqslant F(a, b, 0, \cdots, 0) \\
& =a b\left(a^{2}+b^{2}\right) \\
& =\frac{1}{2}(2 a b)(1-2 a b) \\
& \leqslant \frac{1}{8}
\end{aligned}$$

On the other hand, we have
$$F\left(\frac{1}{2}, \frac{1}{2}, 0, \cdots, 0\right)=\frac{1}{8}$$

Therefore, the maximum value of $F\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ is $\frac{1}{8}$, and thus the constant $c=\frac{1}{8}$.
Equality holds when $x_{1}, x_{2}, \cdots, x_{n}$ have $n-2$ zeros, and the other two are equal (these two equal values can also be zero).