1. (40 points) Let real numbers $a_{1}, a_{2}, \cdots, a_{n} \in[0,2]$, and define $a_{n+1}=a_{1}$, find the maximum value of $\frac{\sum_{i=1}^{n} a_{1}^{2} a_{i+1}+8 n}{\sum_{i=1}^{n} a_{i}^{2}}$.
【Difficulty】
【Knowledge Points】Algebra Module -- Extreme Value Problems
【Reference Answer】When $a_{1}=a_{2}=\cdots=a_{n}=2$, the original expression $=4$ $-10$

points
Next, we prove $\frac{\sum_{i=1}^{n} a_{i}^{2} a_{i+1}+8 n}{\sum_{i=1}^{n} a_{i}^{2}} \geq 4$, i.e., $\sum_{i=1}^{n} a_{i}^{2} a_{i+1}+8 n \geq 4 \sum_{i=1}^{n} a_{i}^{2}$
Since $a_{1}, a_{2}, \cdots, a_{n} \in[0,2]$, we have $a_{i}^{2} a_{i+1} \leq \frac{1}{2} a_{i}^{2} a_{i+1}^{2},\left(a_{i}^{2}-4\right)\left(a_{i+1}^{2}-4\right) \geq 0-20$ points Therefore, $\sum_{i=1}^{n} a_{i}^{2} a_{i+1}+8 n-4 \sum_{i=1}^{n} a_{i}^{2} \geq \frac{1}{2} \sum_{i=1}^{n} a_{i}^{2} a_{i=1}^{2}+8 n-4 \sum_{i=1}^{n} a_{i}^{2}=\frac{1}{2} \sum_{i=1}^{n}\left(a_{i}^{2}-4\right)\left(a_{i+1}^{2}-4\right) \geq 0-30$ points Therefore, $\frac{\sum_{i=1}^{n} a_{i}^{2} a_{i+1}+8 n}{\sum_{i=1}^{n} a_{i}^{2}} \geq 4$, i.e., the maximum value of $\frac{\sum_{i=1}^{n} a_{i}^{2} a_{i+1}+8 n}{\sum_{i=1}^{n} a_{i}^{2}}$ is 4 -40 points