Problem source

A sequence $t_0$, $t_1$, $t_2$, $...$ is said to be 
\textit{strictly increasing}  if $t_{n+1} > t_n$ for all $n\ge{0}\,$.
\begin{questionparts}
\item
The terms of the sequence $x_0\,$, $x_1\,$, $x_2\,$, $\ldots$ satisfy
$$
\ds x_{n+1}=\frac{x_n^2 +6}{5}
$$ for $n\ge{0}\,$. 
Prove that if $x_0 > 3$ then the sequence 
is strictly increasing.
\item The terms of the sequence $y_0\,$, $y_1\,$, $y_2\,$, $\ldots$
satisfy
$$
\ds y_{n+1}= 5-\frac 6 {y_n}
$$ 
for  $n\ge{0}\,$. 
Prove that if $2 < y_0 < 3$ 
then the sequence is strictly 
increasing but that $y_n<3$ for all $n\,$.
\end{questionparts}

Solution source

\begin{questionparts}
\item Suppose $x_n> 3$ then 
\begin{align*}
&& x_{n+1} &= \frac{x_n^2+9-3}{5} \\
&&& \geq \frac{2\sqrt{x_n^2 \cdot 9} - 3}{5} \\
&&&= \frac{6x_n -3}{5} = x_n + \frac{x_n-3}{5} \\
&&&> x_n > 3
\end{align*}

Therefore if $x_i > 3 \Rightarrow x_{i+1} > x_i$ and $x_{i+1} > 3$ so by induction $x_n$ strictly increasing for all $n$.

\item Suppose $2 < y_n < 3$ then

\begin{align*}
&& y_{n+1} &= 5 - \frac6{y_n} \\
&&&< 5 - \frac63 = 3 \\
\\
&& y_{n+1} &= 5 - \frac4{y_n} - \frac{2}{y_n} \\ \\
&&&= y_n + 5 - \frac{2}{y_n} - \left ( y_n + \frac4{y_n} \right) \\
&&&\geq y_n + 5 - \frac{2}{y_n} - 2\sqrt{y_n \frac{4}{y_n}} \\
&&&= y_n + 1 - \frac{2}{y_n}  \\
&&&> y_n
\end{align*}

Therefore if $y_n \in (2,3)$ we have $y_{n+1} \in ( y_n, 3)$ and so $y_n$ is strictly increasing and bounded.
\end{questionparts}