Show that $a_n$ is decreasing 
$a_1 = 2, a_{n+1} = \frac{1}{3 - a_n}$ for $n \ge 2$. Show $a_n$ is decreasing.

First we need to show $a_n > 0$ for all $n$.
$a_2 = 1/2$ and $a_3 = 2/5$ and $a_4 = 5/13$ 
One way we can do this is by showing $3- a_n > 0$. Thus suppose it holds for $n$ then we need to show $\frac{3(3 - a_n) - 1}{3 - a_n} = \frac{8 - 3a_{n}}{3 - a_n} > 0$, which means showing $8 > 3a_{n}$, but I'm having trouble showing it.

Approach $\boldsymbol{1}$:
Note that
$$
\begin{align}
a_{n+1}-a_n
&=\frac1{3-a_n}-a_n\\
&=\frac{a_n^2-3a_n+1}{3-a_n}\\
&=\frac{\left(a_n-\frac32\right)^2-\frac54}{3-a_n}\tag{1}
\end{align}
$$
If $\frac{3-\sqrt5}2\le a_n\le\frac{3+\sqrt5}2$, then $(1)$ is negative and
$$
\frac{3-\sqrt5}2\le\frac1{3-a_n}\le\frac{3+\sqrt5}2\tag{2}
$$
Furthermore, $\frac{3-\sqrt5}2\le2\le\frac{3+\sqrt5}2$.

Approach $\boldsymbol{2}$:
Note that $\frac1{3-x}$ is an increasing function of $x$ on $(0,3)$.  Furthermore, if $a_n\lt3$, then $\frac1{3-a_n}\gt0$.
Since
$$
a_1=2\gt a_2=1\tag{3}
$$
by repeatedly applying $\frac1{3-x}$ to $(3)$, we get that
$$
a_n\gt a_{n+1}\tag{4}
$$
