$\square$ Example 3 Let $n$ be a fixed integer, $n \geqslant 2$.
(1) 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 numbers;
(2) For this constant $c$, determine the conditions under which equality holds.
(40th IMO Problem)
Since the inequality is homogeneous and symmetric, we can assume $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}$ $\geqslant 0$, and $\sum_{i=1}^{n} x_{i}=1$. At this point, we only need to discuss $F\left(x_{1}, x_{2}, \cdots, x_{n}\right)=$ $\sum_{1 \leqslant i < j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) > 0$. In other words, when adjusting $x$ to $x^{\prime}$, the function value $F$ strictly increases. For any $x=\left(x_{1}, x_{2}, \cdots, x_{n}\right)$, after several adjustments, we eventually get
$$\begin{aligned}
F(x) & \leqslant F(a, b, 0, \cdots, 0)=a b\left(a^{2}+b^{2}\right)=\frac{1}{2}(2 a b)(1-2 a b) \\
& \leqslant \frac{1}{8}=F\left(\frac{1}{2}, \frac{1}{2}, 0, \cdots, 0\right)
\end{aligned}$$

Thus, the constant $c=\frac{1}{8}$. The equality holds if and only if two $x_{i}$ are equal (which can be 0), and the rest of the $x_{j}$ are all zero.