2016
Problem - 3302
Let real numbers $a_1$, $a_2$, $\cdots$, $a_{2016}$ satisfy $9a_i\ge 11a_{i+1}^2$ for $i=1, 2,\cdots, 2015$. Define $a_{2017}=a_1$, find the maximum value of $$P=\displaystyle\prod_{i=1}^{2016}(a_i-a_{i+1}^2)$$
Let $a_1=a_2=\cdots=a_{2016}=\frac{1}{2}$. $P=\frac{1}{4^{2016}}$. Hence $max(P)\ge \frac{1}{4^{2016}}$.
Because $9a_i > 11a_{i+1}^2\ge 9a_{i+1}^2$, we found $a_i \ge a_i^2$ holds for every $i=1, 2, \cdots, 2016$.
$$\sum_{i=1}^{2016}(a_i - a_{i+1}^2)=\sum_{i=1}^{2016}(a_i-a_{i}^2)=2016\times\frac{1}{4}-\sum_{i=1}^{2016}\Big(a_i-\frac{1}{2}\Big)^2\le 504$$
By the AGM:
$$\prod_{i=1}^{2016}(a_i-a_{i+1}^2)\le\Big(\frac{1}{2016}\sum_{1}^{2016}(a_i-a_i^2)\Big)^{2016}\le\Big(\frac{1}{2016}\times 504\Big)^{2016} =\frac{1}{4^{2016}}$$
Therefore $\max(P)=\boxed{\frac{1}{4^{2016}}}$