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

holds for all non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n} \geqslant 0$;
b) For this constant $c$, determine the necessary and sufficient conditions for equality to hold.
This article provides a simple solution.
The notation $\sum_{1 \leq i<j \leq \leqslant} f\left(x_{i}, x_{j}\right)$ represents the sum of all terms $f\left(x_{i}, x_{j}\right)$ for which the indices satisfy $1 \leqslant i<j \leqslant n$, and in the following text, it is simply denoted as $\sum f\left(x_{i}, x_{j}\right)$.
Solution: a) When non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ are not all 0, let
$$
\begin{array}{l} 
x= \frac{\sum x_{i} x_{j}}{\left(\sum_{i=1}^{n} x_{i}\right)^{2}}, \\
y=\frac{\sum x_{i} x_{j} x_{k}\left(x_{i}+x_{j}+x_{k}\right)}{\left(\sum_{i=1}^{n} x_{i}\right)^{4}} . \\
\because \sum x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) \\
= \sum x_{i} x_{j}\left[\left(\sum_{k=1}^{n} x_{k}\right)^{2}-2 \sum x_{i} x_{j}-\sum_{\substack{k=1 \\
(k \neq i, j)}}^{n} x_{k}^{2}\right] \\
= \sum x_{i} x_{j}\left(\sum_{k=1}^{n} x_{k}\right)^{2}-2\left(\sum x_{i} x_{j}\right)^{2} \\
-\sum x_{i} x_{j} x_{k}\left(x_{i}+x_{z}+x_{k}\right), \\
\therefore \text { Equation (1) } \Leftrightarrow c \geqslant-2 x^{2}+x-y . \\
\because-2 x^{2}+x-y \leqslant \frac{1}{8}, \text { with equality if and only if }
\end{array}
$$

the necessary and sufficient condition is $x=\frac{1}{4}$ and $y=0$,
$$
\therefore c \geqslant \frac{1}{8}, c_{\min }=\frac{1}{8} \text {, when } x_{1}=x_{2}=\cdots=x_{n}
$$
$=0$ is also valid;
b) When $c=\frac{1}{8}$, the necessary and sufficient condition for $x=\frac{1}{4}$ and $y=0$ is
$$
\begin{array}{l}
\left(\sum_{i=1}^{n} x_{i}\right)^{2}=4 \sum \tilde{x}_{i} x_{j} \text { and } \sum x_{i} x_{j} x_{k}\left(x_{i}\right. \\
\left.+x_{j}+x_{k}\right)=0 \Leftrightarrow \sum_{i=1}^{n} x_{i}^{2}=2 \sum x_{i} x_{j} \text { and } \\
\sum x_{i} x_{j} x_{k}=0 . \\
\because \sum x_{i} x_{j} x_{k}=0 \Leftrightarrow x_{1}, x_{2}, \cdots, x_{n} \text { have at most }
\end{array}
$$
two terms $x_{i}, x_{j}$ not equal to 0, and at this time $\sum_{i=1}^{n} x_{i}^{2}=2 \sum x_{i} x_{j} \Leftrightarrow x_{i}^{2}+x_{j}^{2}=2 x_{i} x_{j} \Leftrightarrow$ $x_{i}=x_{j}$.

Therefore, the necessary and sufficient condition for $x=\frac{1}{4}$ and $y=0$ is that among $x_{1}, x_{2}, \cdots, x_{n}$, there are two equal terms (which can be 0), and the rest are all 0.
Reference
40th IMO Problem Solutions. Middle School Mathematics, 1999(5)