22. Let $n$ be a given positive integer, $n \geqslant 2, a_{1}, a_{2}, \cdots, a_{n} \in(0,1)$, find the maximum value of $\sum_{i=1}^{n} \sqrt[6]{a_{i}\left(1-a_{i}\right)}$, where $a_{n+1}=a_{1}$. (2006 China Western Mathematical Olympiad)
22. By the AM-GM inequality, we have
$$\begin{aligned}
\sqrt[6]{a_{i}\left(1-a_{i+1}\right)}= & 2^{\frac{4}{6}} \sqrt[6]{a_{i}\left(1-a_{i+1}\right) \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}} \leqslant \\
& 2^{\frac{2}{3}} \cdot \frac{1}{6}\left(a_{i}+\left(1-a_{i+1}\right)+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)= \\
& 2^{\frac{2}{3}} \cdot \frac{1}{6}\left(a_{i}-a_{i+1}+3\right)
\end{aligned}$$

Therefore,
$$\begin{array}{c}
\sum_{i=1}^{n} \sqrt[6]{a_{i}\left(1-a_{i+1}\right)} \leqslant 2^{\frac{2}{3}} \cdot \frac{1}{6} \sum_{i=1}^{n}\left(a_{i}-a_{i+1}+3\right)= \\
2^{\frac{2}{3}} \cdot \frac{1}{6} \cdot 3 n=\frac{n}{\sqrt[3]{2}}
\end{array}$$

Equality holds if and only if $a_{1}=a_{2}=\cdots=a_{n}=\frac{1}{2}$, hence the maximum value of $y$ is $\frac{n}{\sqrt[3]{2}}$.