I19 (40-2, Poland) 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 real numbers $x_{1}, x_{2}, \cdots, x_{n}$.
(2) For this constant $c$, determine the necessary and sufficient conditions for equality to hold.
Solution 1 Since the inequality is symmetric, without loss of generality, assume $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n} \geqslant 0$, and since the degrees on both sides of the inequality are equal, we can also assume $\sum_{i=1}^{n} x_{i}=1$. Therefore, we only need to discuss
$$
F\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\sum_{1 \leqslant i0 .
\end{array}
$$

This indicates that when $x$ is adjusted to $x^{\prime}$, the function value $F$ strictly increases. Thus, for any $x=$ $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$, after several adjustments, we can obtain
$$
\begin{aligned}
F(x) & \leqslant F(a, b, 0, \cdots, 0) \\
& =a b\left(a^{2}+b^{2}\right) \\
& =\frac{1}{2}(2 a b)(1-2 a b) \leqslant \frac{1}{8} \\
& =F\left(\frac{1}{2}, \frac{1}{2}, 0, \cdots, 0\right)
\end{aligned}
$$

Therefore, the smallest constant $c=\frac{1}{8}$.
The equality holds if and only if two $x_{i}$ are equal, and the remaining $n-2$ $x_{i}$ are all 0.
Solution 2 When $x_{1}=x_{2}=1, x_{3}=x_{4}=\cdots=x_{n}=0$,
$$
c \geqslant \frac{1}{16} \times 1 \times 1 \times\left(1^{2}+1^{2}\right)=\frac{1}{8} .
$$

Next, we prove the inequality
$$
\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) \leqslant \frac{1}{8}\left(\sum_{1 \leqslant i \leqslant n} x_{i}\right)^{4}
$$

holds for all non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$.
$$
\begin{aligned}
\left(\sum_{1 \leqslant i \leqslant n} x_{i}\right)^{4} & =\left(\sum_{k=1}^{n} x_{k}^{2}+2 \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\right)^{2} \\
& \geqslant 4 \sum_{k=1}^{n} x_{k}^{2} \cdot 2 \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j} \\
& =8 \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j} \sum_{k=1}^{n} x_{k}^{2} \\
& \geqslant 8 \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right)
\end{aligned}
$$

Therefore,
$$
\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) \leqslant \frac{1}{8}\left(\sum_{1 \leqslant i \leqslant n} x_{i}\right)^{4} .
$$

Hence, the minimum value of $c$ is $\frac{1}{8}$, and the equality holds if and only if two $x_{i}$ are equal, and the remaining $n-2$ $x_{i}$ are all 0.