One, (40 points) Let real numbers $a_{1}, a_{2}, \cdots, a_{2016}$ satisfy
$$
\begin{aligned}
9 a_{i} & >11 a_{i+1}^{2}(i=1,2, \cdots, 2015) . \text { Find } \\
& \left(a_{1}-a_{2}^{2}\right)\left(a_{2}-a_{3}^{2}\right) \cdots\left(a_{2015}-a_{2016}^{2}\right)\left(a_{2016}-a_{1}^{2}\right)
\end{aligned}
$$

the maximum value.
Let $P=\prod_{i=1}^{2016}\left(a_{i}-a_{i+1}^{2}\right), a_{2017}=a_{1}$.
From the given, for $i=1,2, \cdots, 2015$, we have $a_{i}-a_{i+1}^{2}>\frac{11}{9} a_{i+1}^{2}-a_{i+1}^{2}>0$.
If $a_{2016}-a_{1}^{2} \leqslant 0$, then $P \leqslant 0$.
Now consider the case where $a_{2016}-a_{1}^{2}>0$.
By the AM-GM inequality, we get
$$
\begin{aligned}
P^{\frac{1}{2016}} & \leqslant \frac{1}{2016} \sum_{i=1}^{2016}\left(a_{i}-a_{i+1}^{2}\right) \\
& =\frac{1}{2016}\left(\sum_{i=1}^{2016} a_{i}-\sum_{i=1}^{2016} a_{i+1}^{2}\right) \\
& =\frac{1}{2016}\left(\sum_{i=1}^{2016} a_{i}-\sum_{i=1}^{2016} a_{i}^{2}\right) \\
& =\frac{1}{2016} \sum_{i=1}^{2016} a_{i}\left(1-a_{i}\right) \\
& \leqslant \frac{1}{2016} \sum_{i=1}^{2016}\left[\frac{a_{i}+\left(1-a_{i}\right)}{2}\right]^{2} \\
& =\frac{1}{2016} \times 2016 \times \frac{1}{4}=\frac{1}{4} \\
& \Rightarrow P \leqslant \frac{1}{4^{2016}} .
\end{aligned}
$$

When $a_{1}=a_{2}=\cdots=a_{2016}=\frac{1}{2}$, the equality holds in the above inequality, and we have $9 a_{i}>11 a_{i+1}^{2}(i=1,2, \cdots, 2015)$, at this point, $P=\frac{1}{4^{2016}}$.
In conclusion, the maximum value sought is $\frac{1}{4^{2016}}$.