【Analysis】Obviously, when $x_{1}=x_{2}=\cdots=x_{n}=0$, inequality (1) always holds.

If $x_{i}(i=1,2, \cdots, n)$ are not all 0, by homogeneity, we can assume $\sum_{i=1}^{n} x_{i}=1$.
When $n=2$, $x_{1}+x_{2}=1$,
the left side of equation (1)
$$\begin{array}{l} =x_{1} x_{2}\left(x_{1}^{2}+x_{2}^{2}\right)=x_{1} x_{2}\left(1-2 x_{1} x_{2}\right) \\ =\frac{1}{2} \times 2 x_{1} x_{2}\left(1-2 x_{1} x_{2}\right) \leqslant \frac{1}{8}, \end{array}$$

with equality holding if and only if $x_{1}=x_{2}=\frac{1}{2}$.
When $n \geqslant 3$, assume $x_{1}+x_{2}<\frac{3}{4}$, and let $x_{3}$, $x_{4}, \cdots, x_{n}$ be fixed.
Let $x_{1}+x_{2}=a$.
Then the left side of equation (1)
$$\begin{array}{l} = \sum_{i=1}^{n}\left(x_{i}^{3} \sum_{j \neq i} x_{j}+x_{i} \sum_{j \neq i} x_{j}^{3}\right) . \\ \text { Let } A=x_{1}^{3}\left(1-x_{1}\right)+x_{2}^{3}\left(1-x_{2}\right) \\ =\left(x_{1}+x_{2}\right)\left[\left(x_{1}+x_{2}\right)^{2}-3 x_{1} x_{2}\right]- \\ {\left[\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}\right]^{2}-2 x_{1}^{2} x_{2}^{2} . } \end{array}$$

Let $x_{1} x_{2}=t$. Then
$$\begin{aligned} A & =a\left(a^{2}-3 t\right)-\left(a^{2}-2 t\right)^{2}-2 t^{2} \\ & =-2 t^{2}+\left(4 a^{2}-3 a\right) t-a^{4} . \end{aligned}$$

Since $a<\frac{3}{4}$, the axis of symmetry is less than 0, and the parabola opens downwards.

Thus, when $x_{1} x_{2}=0$, it has the maximum value. Without loss of generality, assume $x_{1}=0$, at this point, the problem is reduced to an $(n-1)$-variable problem, and so on, it is reduced to a two-variable problem. Therefore, it is known that the equality in inequality (1) holds if and only if two of the variables are equal and the rest are 0.