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)