Prove $a_t \leq \frac{c}{t}$ 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.
