1. Given non-negative real numbers $a_{1}, a_{2}, \cdots, a_{2022}, b_{1}, b_{2}, \cdots, b_{2022}, c_{1}, c_{2}, \cdots, c_{2022}$ satisfy $\sum_{i=1}^{2022} a_{i}=1$, $\sum_{i=1}^{2022} b_{i}^{2}=2, \sum_{i=1}^{2022} c_{i}^{3}=3$. Find the maximum value of $\sum_{i=1}^{2022} a_{i} b_{i+1} c_{i+2}+\sum_{i=3}^{2022} a_{i} b_{i-1} c_{i-2}$.
Given $i \leqslant 0$ or $i \geqslant 2023$, $b_{i}, c_{i}=0$, thus
$$
\sum_{i=1}^{2022} a_{i} b_{i+1} c_{i+2}+\sum_{i=3}^{2022} a_{i} b_{i-1} c_{i-2}
=\sum_{i=1}^{2022} a_{i}\left(b_{i+1} c_{i+2}+b_{i-1} c_{i-2}\right) \leqslant \max _{1 \leqslant i \leqslant 2022}\left\{b_{i+1} c_{i+2}+b_{i-1} c_{i-2}\right\} .
$$

By the AM-GM inequality, we have $b^{2}+b^{2}+b^{2}+\frac{2}{3} c^{3}+\frac{2}{3} c^{3}+1 \geqslant 6 \sqrt[6]{\frac{4}{9}} b c=6 \sqrt[3]{\frac{2}{3}} b c$,
thus $b_{i+1} c_{i+2}+b_{i-1} c_{i-2} \leqslant \frac{1}{6} \sqrt[3]{\frac{3}{2}}\left[3\left(b_{i+1}^{2}+b_{i-1}^{2}\right)+\frac{4}{3}\left(c_{i+2}^{3}+c_{i-2}^{3}\right)+2\right] \leqslant \frac{1}{6} \sqrt[3]{\frac{3}{2}} \cdot 12=\sqrt[3]{12}$, equality holds when $a_{3}=1, b_{2}=b_{4}=1, c_{1}=c_{5}=\sqrt[3]{\frac{3}{2}}$, and the rest are 0. Therefore, the maximum value of $\sum_{i=1}^{2022} a_{i} b_{i+1} c_{i+2}+\sum_{i=3}^{2022} a_{i} b_{i-1} c_{i-2}$ is $\sqrt[3]{12}$.
Note: Since the sum of 2 $b_{i}^{2}$ is 2 and the sum of 2 $c_{i}^{3}$ is 3 when the equality holds, the condition for equality in the AM-GM inequality is $b=1, c=\sqrt[3]{\frac{3}{2}}$.