Example 6 Let $n$ be a fixed integer, $n \geqslant 2$. Determine the smallest positive number $c$ such that the inequality
$$
\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) \leqslant c\left(\sum_{i=1}^{n} x_{i}\right)^{4}
$$

holds for all non-negative real numbers $x_{i}(i=1,2, \cdots, n)$.
(40th IMO)
Explanation: When $x_{i}=0(i=1,2, \cdots, n)$, it holds for all $c$.
When $x_{i}$ are not all 0, the problem transforms into
$$
f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\frac{\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right)}{\left(\sum_{i=1}^{n} x_{i}\right)^{4}} \leqslant c
$$

always holds, i.e.,
$$
f\left(x_{1}, x_{2}, \cdots, x_{n}\right)_{\max } \leqslant c .
$$

Since $f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ is a homogeneous symmetric function, and without loss of generality, let $\sum_{i=1}^{n} x_{i}=1$, then
$$
f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\sum_{i=1}^{n}\left(x_{i}^{3}-x_{i}^{4}\right) .
$$

Without loss of generality, assume $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}$. If $x_{k+1}=$ $x_{k+2}=\cdots=x_{n}=0$, adjust
$$
(x_{1}, x_{2}, \cdots, x_{k-1}, x_{k}, \underbrace{0,0, \cdots, 0}_{(n-k) \uparrow})
$$

to
$$
(x_{1}, x_{2}, \cdots, x_{k-1}+x_{k}, \underbrace{0,0, \cdots, 0}_{(n-k+1) \uparrow}) \text {, }
$$

then
$$
f(x_{1}, x_{2}, \cdots, x_{k-1}+x_{k}, \underbrace{0,0, \cdots, 0}_{(n-k+1) \uparrow})-
$$
$$
\begin{array}{l} 
f(x_{1}, x_{2}, \cdots, x_{k-1}, x_{k}, \underbrace{0,0, \cdots, 0}_{(n-k) \uparrow}) \\
=\left(x_{k-1}+x_{k}\right)^{3}-\left(x_{k-1}+x_{k}\right)^{4}- \\
\left(x_{k-1}^{3}-x_{k-1}^{4}+x_{k}^{3}-x_{k}^{4}\right) . \\
= 3 x_{k-1}^{2} x_{k}+3 x_{k-1} x_{k}^{2}-\left(4 x_{k-1}^{3} x_{k}+\right. \\
\left.6 x_{k-1}^{2} x_{k}^{2}+4 x_{k-1} x_{k}^{3}\right) \\
= x_{k-1} x_{k}\left[2\left(x_{k-1}+x_{k}\right)-3\left(x_{k-1}+x_{k}\right)^{2}+\right. \\
\left.\left(x_{k-1}-x_{k-1}^{2}\right)+\left(x_{k}-x_{k}^{2}\right)\right] .
\end{array}
$$

If $k-1 \geqslant 2$, then $x_{k-1}+x_{k} \leqslant \frac{2}{3}$.
Also, $x_{k}-x_{k}^{2} \geqslant 0, x_{k-1}-x_{k-1}^{2} \geqslant 0$, so
$$
\begin{array}{l}
f(x_{1}, x_{2}, \cdots, x_{k-1}+x_{k}, \underbrace{0,0, \cdots, 0}_{(n-k+1) \uparrow}) \\
\geqslant f(x_{1}, x_{2}, \cdots, x_{k-1}, x_{k}, \underbrace{0,0, \cdots, 0}_{(n-k) \uparrow}) .
\end{array}
$$

After one adjustment, $f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ does not decrease, hence after a finite number of adjustments, we have
$$
\begin{array}{l}
f\left(x_{1}, x_{2}, \cdots, x_{n}\right) \leqslant f(x_{1}, x_{2}, \underbrace{0,0, \cdots, 0}_{(n-2) \uparrow})) \\
=x_{1}^{3}+x_{2}^{3}-x_{1}^{4}-x_{2}^{4} .
\end{array}
$$

Since $x_{1}+x_{2}=1$, we have
$$
\begin{array}{l}
f(x_{1}, x_{2}, \underbrace{0, \cdots, \cdots, 0}_{(n-2) \uparrow})=x_{1} x_{2}\left(1-2 x_{1} x_{2}\right) \leqslant \frac{1}{8} . \\
\text { When } x_{1}=x_{2}=\frac{1}{2}, x_{i}=0(i=3,4, \cdots, n)
\end{array}
$$

then
$$
f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\frac{1}{8} .
$$

Therefore, $f\left(x_{1}, x_{2}, \cdots, x_{n}\right)_{\text {max }}=\frac{1}{8}$.
Thus, $c \geqslant \frac{1}{8}$, i.e., $c_{\text {min }}=\frac{1}{8}$.
Note: The step-by-step adjustment yields
$$
\begin{array}{l}
f(x_{1}, x_{2}, \cdots, x_{k-1}+x_{k}, \underbrace{0,0, \cdots, 0}_{(n-k+1) \uparrow}) \\
\geqslant f(x_{1}, x_{2}, \cdots, x_{k-1}, x_{k}, \underbrace{0,0, \cdots, 0}_{(n-k) \uparrow}),
\end{array}
$$

reducing the number of variables step by step is the key to this solution.