Let $n$ be a given positive integer, $n \geqslant 2, a_{1}, a_{2}, \cdots, a_{n} \in(0,1)$. Find
$$
\sum_{i=1}^{n} \sqrt[6]{a_{i}\left(1-a_{i+1}\right)}
$$

the maximum value, where $a_{n+1}=a_{1}$. (Supplied by Hua-Wei Zhu)
By the AM-GM inequality, we have
$$
\begin{aligned}
& \sqrt[6]{a_{i}\left(1-a_{i+1}\right)} \\
= & 2^{\frac{1}{6}} \sqrt[6]{a_{i}\left(1-a_{i+1}\right) \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}} \\
\leqslant & 2^{\frac{2}{3}} \cdot \frac{1}{6}\left(a_{i}+1-a_{i+1}+2\right) \\
= & 2^{\frac{2}{3}} \cdot \frac{1}{6}\left(a_{i}-a_{i+1}+3\right),
\end{aligned}
$$

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

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