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.
 A: If we have that $x_k\geq 3$, then we have that
$$x_{k+1}=\frac{x_k^2}{2}-1\geq \frac{3x_k}{2}-1=x_k+\frac{x_k}{2}-1\geq x_k+\frac{3}{2}-1=x_k+\frac{1}{2}$$
So
$$x_2 \geq 3 + \frac{1}{2}$$
$$x_3 \geq 3 + \frac{1}{2} + \frac{1}{2}$$
$$...$$
$$x_n \geq 3 + \frac{n-1}{2}$$
And it's enough to see that $(x_n)_{n \geq 1}$ is increasing and unbounded from above.
A: Since the sequence is increasing just use the fact that $x_n > x_1 = 3$ for all $n$.
To show the sequence is not bounded above, show by induction that $x_n > n$ for all $n$.
A: $x_{n+1} - x_n = \dfrac{x_n^2}{2}-1 - x_n= \dfrac{x_n^2 - 2x_n - 1}{2}=\dfrac{(x_n-1)^2-2}{2}$. 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 $3$.
A: You can start with the definition of lower bound. A real number α is called a lower bound for S if α ≤ x for all x ∈ S. Saying that all elements are not less than 3 is Ok. That is why it is bounded below.
To show that it is unbounded below you can use 
Completeness Axiom:Any nonempty subset of R that is bounded above has a least upper bound.
Assuming that sequence is bounded above and then having a contradiction of the definition of sup
A: Since the sequence is strictly increasing, it is sufficient to say that it will never go below it's first value $x_1$. 
Your induction is mostly correct. The implication $x_{n+1}>x_n \Rightarrow x_{n+1}^2>x_n^2$ requires both values to be positive which is included in the initial assumption (but I would mention it).
For the upper bound you could show by another induction that the sequence always increases at least by $0.5$.
