Prove that if $2 < y_0 < 3$ then the sequence is strictly increasing but that $y_n<3$ for all $n\,$ 
A sequence $t_0$, $t_1$, $t_2$, $...$ is said to be strictly
increasing  if $t_{n+1} > t_n$ for all $n\ge{0}\,$.
The terms of the sequence $y_0\,$,$y_1\,$, $y_2\,$, $\ldots$
satisfy $$ y_{n+1}= 5-\frac 6 {y_n} $$  for  $n\ge{0}\,$.  Prove that
if $2 < y_0 < 3$  then the sequence is strictly  increasing but that
$y_n<3$ for all $n\,$.

Hello. I am not very good at these types of questions and require some assistance.
Workings: I have started off with $$y_{n+1} - y_n > 0 \\ \implies 5-\frac{6}{y_n} - y_n>0 \\\implies \frac{-y_n^2+5y_n-6}{y_n}>0  $$
So, from this, we see that when $2<y_n<3$ the sequence is increasing. Additionally, it's increasing when $y_n<0$. However, I am not sure why the question didn't include that part? Anyway, I don't know how to make any progress from $2<y_n<3$.
 A: Let us prove by induction that $2 < y_{n} < 3$. When $n = 0$, the relation clearly holds.
Suppose then it holds for $n$. We shall prove it holds for $n + 1$ as well:
\begin{align*}
2 < y_{n} < 3 & \Rightarrow \frac{1}{3} < \frac{1}{y_{n}} < \frac{1}{2}\\\\
& \Rightarrow -\frac{1}{2} < -\frac{1}{y_{n}} < -\frac{1}{3}\\\\
& \Rightarrow -3 < -\frac{6}{y_{n}} < -2\\\\
& \Rightarrow 2 < 5 - \frac{6}{y_{n}} < 3\\\\
& \Rightarrow 2 < y_{n+1} < 3
\end{align*}
and we are done.
Hopefully this helps!
A: A different solution than the one provided by Atila Correia, following the same thought process you initially did, is the following:
The problem states that
$2 < y_0 < 3 \tag{1}$
It's also known that
$y_{n+1} = 5 - \frac{6}{y_n} \tag{2}$
Starting from the fact that the sequence is strictly increasing (what we want to prove) we have:
\begin{equation}
    y_{n+1} - y_{n} = (5 - \frac{6}{y_n}) - y_n
    = \frac{-y_{n}^2 + 5y_n - 6}{y_n} > 0 \tag{3}
\end{equation}
Thus,
\begin{cases}
  -y_n^2 + 5y_n - 6 > 0\\
  y_n > 0
  \tag{4}
\end{cases}
or
\begin{cases}
  -y_n^2 + 5y_n - 6 < 0\\
  y_n < 0
  \tag{5}
\end{cases}
Based on $(1)$, $y_n$ can't be negative, then $(5)$ is impossible. From $(4)$ we have
\begin{equation}
  -y_n^2 + 5y_n - 6 > 0 \Rightarrow
  \begin{cases}
    y_n > 2\\
    y_n < 3
  \end{cases}
  \tag{6}
\end{equation}
Thus $y_n < 3 \ \forall\ n \geq 0$, if the sequence is strictly increasing.
For the cases that the sequence is not strictly increasing we should have:
\begin{cases}
  -y_n^2 + 5y_n - 6 > 0\\
  y_n < 0
  \tag{7}
\end{cases}
or
\begin{cases}
  -y_n^2 + 5y_n - 6 < 0\\
  y_n > 0
  \tag{8}
\end{cases}
From $(1)$ we have that $y_n$ can't be negative for all $n$, so $(7)$ doesn't hold.
Let's call $f$ the $-y_n^2 + 5y_n - 6$ curve. For $y_n > 0$, $f$ behaves as follows:
\begin{cases}
  0 < y_n < 2 & \implies f < 0\\
  2 < y_n < 3 & \implies f > 0\\
  3 < y_n & \implies f < 0
  \tag{9}
\end{cases}
So $(8)$ only holds if $y_n$ is not in the $(2, 3)$ interval. That doesn't satisfy $(1)$, meaning $(8)$ also doesn't hold. This means the sequence is necessarily strictly increasing.
