5. (40th IMO Problem) $n$ is a given integer, $n \geqslant 2$, 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 $x_{1}, x_{2}, \cdots, x_{n}$.
5. Solution If $x_{1}, x_{2}, \cdots, x_{n}$ are all 0, $c$ can be any real number.

If $x_{1}, x_{2}, \cdots, x_{n}$ have at least one positive number. Without loss of generality, assume $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}$. Since the inequality is homogeneous, we can set $\sum_{i=1}^{n} x_{i}=1$. Let $F\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\sum_{1 \leqslant i0$. This indicates that after adjusting $x$ to $x^{\prime}$, $F(x)$ strictly increases.
After several adjustments for $x=\left(x_{1}, x_{2}, \cdots, x_{n}\right)$, we eventually get $F(x) \leqslant F\left(x_{1}, x_{2}, 0, \cdots, 0\right)=x_{1} x_{2}\left(x_{1}^{2}+\right.$ $\left.x_{2}^{2}\right)=\frac{1}{2}\left(1-2 x_{1} x_{2}\right) \cdot 2 x_{1} x_{2} \leqslant \frac{1}{8}$. Therefore, $c \geqslant \frac{1}{8}$. In summary, $c_{\min }=\frac{1}{8}$.