We have the recursive equation $$a_t^2 = (1 - a_t)a_{t-1}^2$$ $$a_0 = 1$$
Prove that there exists constant $c$ such that $a_t \leq \frac{c}{t}$, and find the smallest such $c$.
I was able to get that $a_t^2 = (1 - a_t)(1 - a_{t-1})\dots(1 - a_1)$, but I don't know how to proceed after that, and I'm not sure if I'm going in the right direction.
It turn out that $c \ge 2$ will work.
$a_t^2 = (1 - a_t)a_{t-1}^2 $.
$a_1^2 = 1-a_1$ so $a_1 =\dfrac{-1\pm\sqrt{5}}{2} =\dfrac{-1+\sqrt{5}}{2} $.
$a_t^2+a_ta_{t-1}^2-a_{t-1}^2 = 0 $ so
$\begin{array}\\ a_t &=\dfrac{-a_{t-1}^2+\sqrt{a_{t-1}^4+4a_{t-1}^2}}{2}\\ &=\dfrac{-a_{t-1}^2+a_{t-1}\sqrt{a_{t-1}^2+4}}{2}\\ &=a_{t-1}\dfrac{-a_{t-1}+\sqrt{a_{t-1}^2+4}}{2}\\ &=a_{t-1}\dfrac{-a_{t-1}+\sqrt{a_{t-1}^2+4}}{2}\dfrac{a_{t-1}+\sqrt{a_{t-1}^2+4}}{a_{t-1}+\sqrt{a_{t-1}^2+4}}\\ &=a_{t-1}\dfrac{2}{a_{t-1}+\sqrt{a_{t-1}^2+4}}\\ &=\dfrac{2}{1+\sqrt{1+4/a_{t-1}^2}}\\ \text{if} &a_{t-1} \le \dfrac{c}{t-1}\\ a_t &\le\dfrac{2}{1+\sqrt{1+4/(c/(t-1))^2}}\\ &=\dfrac{2}{1+\sqrt{1+4(t-1)^2/c^2}}\\ &\lt\dfrac{2}{1+\sqrt{4(t-1)^2/c^2}}\\ &=\dfrac{2}{1+2(t-1)/c}\\ \end{array} $
so we want $\dfrac{2}{1+2(t-1)/c} \le \dfrac{c}{t} $ or $\dfrac{2c}{c+2(t-1)} \le \dfrac{c}{t} $ or $2t \le c+2(t-1) =2t+c-2 $ or $c \ge 2$.
Therefore $c \ge 2$ will work.