（40分）设实数 $a_{1},a_{2},\cdots,a_{2016}$ 满足 $9a_{i}>11a_{i+1}^{2},i=1,2,\cdots,2015$．求 $(a_{1}-a_{2}^{2}) (a_{2}-a_{3}^{2})\cdots (a_{2015}-a_{2016})^{2}(a_{2016}-a_{1}^{2})$ 的最大值．
【难度】
【出处】
无
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【答案】
$\dfrac{1}{4^{2016}}$
【解析】
令 $p=(a_{1}-a_{2}^{2})\cdot (a_{2}-a_{3}^{2})\cdots (a_{2015}-a_{2016}^{2})(a_{2016}-a_{1}^{2})$．由已知得，对 $i=1,2,\cdots,2015$，均有\[a_{i}-a_{i+1}^{2}>\dfrac{11}{9}a_{i+1}^{2}-a_{i+1}^{2}\geqslant 0.\]若 $a_{2016}-a_{1}^{2}\leqslant 0$，则 $S\leqslant 0$．
以下考虑 $a_{2016}-a_{1}^{2}>0$ 的情况．约定 $a_{2017}=a_{1}$．由平均不等式得\[\begin{split}P^{frac{1}{2016}}&\leqslant \dfrac{1}{2016}\sum\limits_{i=1}^{2016}(a_{i}-a_{i+1}^{2})\\&=\dfrac{1}{2016}\left(\sum\limits_{i=1}^{2016}a_{i}-\sum\limits_{i=1}^{2016}a_{i}^{2}\right)\\&=\dfrac{1}{2016}\sum\limits_{i=1}^{2016}a_{i}(1-a_{i})\\&\leqslant \dfrac{1}{2016}\sum\limits _{i=1}^{2016}\left(\dfrac{a_{i}+(1-a_{i})}{2}\right)^{2}\\&=\dfrac{1}{2016}\cdot 2016\cdot \dfrac{1}{4}=\dfrac{1}{4},\end{split}\]所以 $P\leqslant \dfrac{1}{4^{2016}}$，当 $a_{1}=a_{2}=\cdots =a_{2016}=\dfrac{1}{2}$ 时，上述不等式等号成立，且有\[9a_{i}>11a_{i+1}^{2},i=1,2,\cdots 2015,\]此时 $P=\dfrac{1}{4^{2016}}$．
以下考虑 $a_{2016}-a_{1}^{2}>0$ 的情况．约定 $a_{2017}=a_{1}$．由平均不等式得\[\begin{split}P^{frac{1}{2016}}&\leqslant \dfrac{1}{2016}\sum\limits_{i=1}^{2016}(a_{i}-a_{i+1}^{2})\\&=\dfrac{1}{2016}\left(\sum\limits_{i=1}^{2016}a_{i}-\sum\limits_{i=1}^{2016}a_{i}^{2}\right)\\&=\dfrac{1}{2016}\sum\limits_{i=1}^{2016}a_{i}(1-a_{i})\\&\leqslant \dfrac{1}{2016}\sum\limits _{i=1}^{2016}\left(\dfrac{a_{i}+(1-a_{i})}{2}\right)^{2}\\&=\dfrac{1}{2016}\cdot 2016\cdot \dfrac{1}{4}=\dfrac{1}{4},\end{split}\]所以 $P\leqslant \dfrac{1}{4^{2016}}$，当 $a_{1}=a_{2}=\cdots =a_{2016}=\dfrac{1}{2}$ 时，上述不等式等号成立，且有\[9a_{i}>11a_{i+1}^{2},i=1,2,\cdots 2015,\]此时 $P=\dfrac{1}{4^{2016}}$．
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