For example, $6.8 x_{i} \geqslant 0, i=1,2, \cdots, n, n \geqslant 2$, find the smallest $C$ such that
$$C\left(\sum_{i=1}^{n} x_{i}\right)^{4} \geqslant \sum_{1<i<j<n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right)$$
On one hand, taking $x_{1}=x_{2}=1, x_{3}=x_{4}=\cdots=x_{n}$, we get $C \geqslant \frac{1}{8}$.
On the other hand, by the AM - GM inequality, we have
$$\begin{aligned}
\left(\sum_{i=1}^{n} x_{i}\right)^{4}= & \left(\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\right)^{2} \geqslant 4\left(\sum_{i=1}^{n} x_{i}^{2}\right)\left(2 \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\right)= \\
& 8\left(\sum_{i=1}^{n} x_{i}^{2}\right) \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j} \geqslant 8 \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right)
\end{aligned}$$

Thus, $C_{\text {min }}=\frac{1}{8}$.
Note: The general solution to this problem involves adjustment, and the above proof is quite ingenious.
Example 6.9: Given $a, b, c \geqslant 0$, prove that
$$a^{2} b^{2}(a-b)^{2}+b^{2} c^{2}(b-c)^{2}+c^{2} a^{2}(c-a)^{2} \geqslant 2 a b c(a-b)(b-c)(c-a)$$

Proof: By the AM - GM inequality, we have
$$\begin{aligned}
\sum\left[a^{2} b^{2}(a-b)^{2}\right]= & \frac{1}{2} \sum a^{2}\left[b^{2}(a-b)^{2}+c^{2}(a-c)^{2}\right] \geqslant \\
& a b c \sum a|(a-b)(a-c)|
\end{aligned}$$

Clearly,
$$2 a b c(a-b)(b-c)(c-a) \leqslant 2 a b c|(a-b)(b-c)(c-a)|$$

Therefore, it suffices to prove
$$\sum a|(a-b)(a-c)| \geqslant 2|(a-b)(b-c)(c-a)| \Leftrightarrow \sum\left|\frac{a}{b-c}\right| \geqslant 2$$

Without loss of generality, assume $a \geqslant b \geqslant c$, then we have
$$\text { LHS } \geqslant \frac{a}{b-c}+\frac{b}{a-c} \geqslant \frac{a}{b}+\frac{b}{a} \geqslant 2$$

The proposition is proved!