Prove convergence of $x_{n+1} = \frac{x_n}{2} + \frac{2}{x_n}$ Given $x_1= 1$ and  $$x_{n+1} = \frac{x_n}{2} + \frac{2}{x_n}$$ for all $n\in\mathbb{N}, n\geq 1$.  Prove that the sequence $\{x_n\}$ converges and find its limit.
The hint given says that taking the limit of both sides is only valid if you know the limit exists already. 
Part a of the question determines that $ab \leq \frac{a^2 + b^2}{2}$.
I assume we try to use Cauchy along with a trick from part (a)? However, I have no clue how to go about that.  Thank you!
 A: I restate the question in the general setting.
QUESTION
Let $\{x_n\}$ be the sequence given by the following formula:
$$
x_{n+1}=\frac{x_n}{2}+\frac{2}{x_n} (n\in\mathbb{N}, n\geq 1)
$$
with $x_1\ne 0$. Prove that $\{x_n\}$ converges.
SOLUTION.
We consider two cases of $x_1$.
Case 1. $x_1> 0$
By induction we have $x_n> 0$ for all $n\in\mathbb{N}$. Using AM-GM inequality we get $x_n\geq 2$ for all $n\in\mathbb{N}, n>1$. Moreover,
$$
x_{n+1}-x_n=\frac{2}{x_n}-\frac{x_n}{2}=\frac{4-x_n^2}{2x_n}\leq 0.
$$
Since $\{x_n\}$ is a decreasing sequence and lower bounded by $0$, it converges.
Case 2. $x_1< 0$
By induction we have $x_n< 0$ for all $n\in\mathbb{N}$. Using AM-GM inequality we get $x_n\leq -2$ for all $n\in\mathbb{N}, n>1$. Moreover,
$$
x_{n+1}-x_n=\frac{2}{x_n}-\frac{x_n}{2}=\frac{4-x_n^2}{2x_n}\geq 0.
$$
Since $\{x_n\}$ is an increasing sequence and upper bounded by $0$, it converges.
NOTE We make clear how to use AM-GM inequality in both cases:
$$
x_{n+1}^2=\left(\frac{x_n}{2}+\frac{2}{x_n}\right)^2=\frac{x_n^2}{4}+\frac{4}{x_n^2}+2\geq 2\sqrt{\frac{x_n^2}{4}.\frac{4}{x_n^2}}+2=4.
$$
Hence, $|x_{n+1}|\geq 2$ for all $n\in\mathbb{N}, n>1$.
A: The first thing I see here is that your sequence resembles the "Babylonian method" to find square roots.
That is the Babylonians figured out that : $x_{n+1} = \frac 12 (\frac {N}{x_n} + x_n)$ is a sequence that converges to $\sqrt {N}$
In our case 
$x_{n+1} = \frac 12 (\frac {4}{x_n} + x_n)$
We can guess that $\{x\}$ is converging to $2.$
How are we going to prove it?
Show that if $x_n > 2$ then $2<x_{n+1}< x_n$
$x_0 = 1 \implies x_1 = \frac{5}{2}$ 
For all $n>1$ the sequence monotonic (decreasing) and bounded (below) and therefore convergent.
A: as pointed out by Doug M, $x_1 = 5/2$ which is above $2$
Once $x > 2,$ we have $$ 0 < \frac{x-2}{2x} < \frac{1}{2} $$
The interesting thing is how $x-2$ shrinks under your mapping:
$$ \frac{x}{2} + \frac{2}{x} - 2 = \left(\frac{x-2}{2x} \right) (x-2) , $$
so with $x > 2$ we get
$$ \frac{x}{2} + \frac{2}{x} - 2 < \left(\frac{1}{2} \right) (x-2) , $$
with $n \geq 1$ we have
$$ (x_{n+1} - 2 ) < \frac{1}{2} (x_n - 2)  $$
You don't need the next thing: there is actually quadratic convergence.
