[Example 3.3.9] Let $n$ be a fixed integer, $n \geqslant 2$
(I) Determine the smallest constant $c$ such that the inequality
$$
\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) \leqslant c\left(\sum_{i=1}^{n} x_{i}\right)^{4} ;
$$
(II) For this constant $c$, determine the necessary and sufficient conditions for equality to hold.
Due to the inequality being 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, the given inequality transforms into
$$
\sum_{1 \leqslant i0 \text {, }
$$

which means
$$
\begin{aligned}
F\left(x^{\prime}\right)-F(x) & =x_{k} x_{k+1}\left\{\left(x_{k}+x_{k+1}\right)\left[3-4\left(x_{k}+x_{k+1}\right)\right]+2 x_{k} x_{k+1}\right\} \\
& >0 .
\end{aligned}
$$

Therefore,
$$
F\left(x^{\prime}\right)-F(x)>0 .
$$

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

Equality holds if and only if $a=b=\frac{1}{2}$.
That is, $F\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\left(\frac{1}{2}, \frac{1}{2}, 0, \cdots, 0\right)$.
Thus, the smallest constant $c=\frac{1}{8}$.
Equality holds if and only if two of the $x_{i}$ are equal, and the rest of the $x_{i}$ are 0.