1. Let $n(n \geqslant 2)$ be a given positive integer, and $a_{1}, a_{2}$, $\cdots, a_{n} \in(0,1)$. Find the maximum value of $\sum_{i=1}^{n} \sqrt[6]{a_{i}\left(1-a_{i+1}\right)}$, where $a_{n+1}=a_{1}$.
(Provided by Hua-Wei Zhu)
1. By the arithmetic-geometric mean inequality, we have
$$
\begin{array}{l}
\sqrt[6]{a_{i}\left(1-a_{i+1}\right)} \\
=2^{\frac{4}{6}} \sqrt[6]{a_{i}\left(1-a_{i+1}\right) \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}} \\
\leqslant 2^{\frac{2}{3}} \times \frac{1}{6}\left(a_{i}+1-a_{i+1}+2\right) \\
=2^{\frac{2}{3}} \times \frac{1}{6}\left(a_{i}-a_{i+1}+3\right),
\end{array}
$$

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

The equality holds if and only if $a_{1}=a_{2}=\cdots=a_{n}=\frac{1}{2}$. Hence, the maximum value of $y$ is $\frac{\sqrt[3]{4} n}{2}$.