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 \(x_0\,\), \(x_1\,\), \(x_2\,\), \(\ldots\) 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.
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 \(n\,\).
Solution:
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 \(n\).