Example 8 Given an integer $n \geqslant 2$, try to determine the smallest constant $c$, such that for any non-negative real numbers $x_{i}(1 \leqslant i \leqslant n)$, we have $\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) \leqslant c\left(\sum_{i=1}^{n} x_{i}\right)^{4}$. And for this constant $c$, determine the necessary and sufficient conditions for equality to hold.
Let $S_{k}=x_{1}^{k}+x_{2}^{k}+\cdots+x_{n}^{k}, k=1,2,3,4$. Using Newton's formula $S_{4}-\sigma_{1} S_{3}+\sigma_{2} S_{2} -\sigma_{3} S_{1}+4 \sigma_{4}=0$, express $\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right)$ as a polynomial in terms of $\sigma_{1}, \sigma_{2}, \sigma_{3}, \sigma_{4}$, and determine the smallest constant $c$.
Solution: Let $S_{k}=x_{1}^{k}+x_{2}^{k}+\cdots+x_{n}^{k}, k=1,2,3,4$. According to Newton's formula, we have
$S_{4}-\sigma_{1} S_{3}+\sigma_{2} S_{2}-\sigma_{3} S_{1}+4 \sigma_{4}=0$.
Assume $S_{1}=\sigma_{1}=1$, then
$$
\begin{array}{l}
F\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right)=\sum_{i=1}^{n}\left(x_{i}^{3} \sum_{j \neq i} x_{j}\right) \\
=\sum_{i=1}^{n} x_{1}\left(1-x_{i}\right)=S_{3}-S_{4}=\sigma_{2} S_{2}-\sigma_{3} S_{1}+4 \sigma_{4}=\sigma_{2}\left(1-2 \sigma_{2}\right)+4 \sigma_{4}-\sigma_{3} .
\end{array}
$$

Since $\sigma_{2}\left(1-2 \sigma_{2}\right)=\frac{1}{2} \cdot 2 \sigma_{2}\left(1-2 \sigma_{2}\right) \leqslant \frac{1}{2} \cdot\left(\frac{2 \sigma_{2}+1-2 \sigma_{2}}{2}\right)^{2}=\frac{1}{8}$,
and $4 \sigma_{4}-\sigma_{3}=\sum_{1 \leqslant i_{1}<i_{2}<i_{3} \leqslant n} x_{i_{1}} x_{i_{2}} x_{i_{3}}\left(\sum_{i \notin\left\{i_{1}, i_{2}, i_{3}\right\}} x_{i}\right)-\sum_{1 \leqslant i_{1}<i_{2}<i_{3} \leqslant n} x_{i_{1}} x_{i_{2}} x_{i_{3}}$
$$
\begin{array}{l}
=\sum_{1 \leqslant i_{1}<i_{2}<i_{3} \leqslant n} x_{i_{1}} x_{i_{2}} x_{i_{3}}\left(\sum_{i \notin\left\{i_{1}, i_{2}, i_{3}\right\}} x_{i}-1\right) \\
=-\sum_{1 \leqslant i_{1}<i_{2}<i_{3} \leqslant n} x_{i_{1}} x_{i_{2}} x_{i_{3}}\left(x_{i_{1}}+x_{i_{2}}+x_{i_{3}}\right)
\end{array}
$$
$$
\leqslant 0,
$$

Therefore, $F\left(x_{1}, x_{2}, \cdots, x_{n}\right) \leqslant \frac{1}{8}$, hence $c=\frac{1}{8}$. The equality holds if and only if $\sigma_{2}=\frac{1}{4}$, and at least $n-2$ of $x_{i}(1 \leqslant i \leqslant n)$ are zero, i.e., exactly $n-2$ of $x_{i}(1 \leqslant i \leqslant n)$ are zero, and the other two are equal.