<s>[INST] Show $x_{n+1} = {1\over 2}x_n^2 - 1$ is bounded below and unbounded above and $x_n$ is increasing. 
Let: 
  $$
\begin{cases}
x_{n+1} = {1\over 2}x_n^2 - 1\\
x_1 = 3\\
n\in \mathbb N
\end{cases}
$$
  Show that the sequence $x_n$ is bounded only below and is increasing.

I've started with the following:
$$
x_1 = 3 \\
x_2 = 3.5 \\
x_2 > x_1
$$
Suppose $x_{n+1} > x_n$. Consider the following equation:
$$
x_{n+2} - x_{n+1} = \frac{1}{2}x_{n+1}^2 - \frac{1}{2}x_n^2
$$
By initial assumption:
$$
x_{n+1} > x_n \implies x_{n+1}^2 > x_n^2 \implies \frac{1}{2} x_{n+1}^2 > \frac{1}{2} x_n^2 \implies x_{n+2} > x_{n+1}
$$
Thus $x_n$ sequence is increasing. 
But now how do I show the lower bound exist and upper does not? Intuitively for the sequence to be unbounded we need the following condition to be satisfied:
$$
{1\over 2}x_n^2 > 1 \iff x_n > \sqrt2
$$
I'm kindly asking to verify whether i've correctly shown that $x_n$ is monotonically increasing and help with showing the bounds.
 [/INST] $x_{n+1} - x_n = \dfrac{x_n^2}{2}-1 - x_n= \dfrac{x_n^2 - 2x_n - 1}{2}=\dfrac{(x_n-1)^2-2}{2}$. Thus we need to show: $x_n \ge 3, \forall n \ge 1$. We show this by induction on $n\ge 1$. We have $x_1 = 3$, assume $x_n \ge 3$, we have $x_{n+1} = \dfrac{x_n^2}{2} - 1 \ge \dfrac{3^2}{2}-1= 4.5-1 = 3.5 > 3$. Thus $x_{n+1} - x_n \ge \dfrac{(3-1)^2-2}{2}=1 > 0$. Thus $x_n$ is a strictly increasing sequence, and clearly bounded below by $3$.
</s>