Showing that $x_{n+1} = 1/2(x_n + 2/x_n)$ is a decreasing sequence. I need to show that $x_{n+1} = 1/2(x_n + 2/x_n)$ is a decreasing sequence, where $x_1 = 2$ and n = 1,2,3,....
I tried to show this with induction, where since $x_1= 2$ then $x_2 = 1.5$, hence the base case satisfies.
Assuming that $x_n \geq x_{n+1}$ , I need to prove that $x_{n+1} \geq x_{n+2}$.
I started with the statement $x_n \geq x_{n+1} \Rightarrow \frac{1}{x_n} \leq \frac{1}{x_{n+1}}$
Also, $x_n \geq x_{n+1}$ then $\frac{1}{2}(x_n + \frac{2}{x_{n+1}}) \geq \frac{1}{2}(x_{n+1} + \frac{2}{x_{n+1}})$
$= \frac{1}{2}(x_n + \frac{2}{x_{n+1}} - \frac{2}{x_n} + \frac{2}{x_n})  \geq x_{n+2} $
$= x_{n+1} - \frac{1}{x_n} + \frac{1}{x_{n+1}}  \geq x_{n+2} $
since $- \frac{1}{x_n} + \frac{1}{x_{n+1}} \geq 0$
then I cannot say that  $ x_{n+1} \geq x_{n+2} $
How do I proceed forward?
Edit:  Since $x_1 = 2$, hence $1 \leq x_1 \leq 2$ holds for base case.
If I assume that $1 \leq x_n \leq 2 $ and show that $1 \leq x_{n+1} \leq 2 $ then by induction we can say that $1 \leq x_n \leq 2 \:\forall n$
I have shown this as:
$x_{n+1} = \frac{x_n}{2} + \frac{1}{x_n}, \: now \: since\:1 \leq x_n\leq2\:\Rightarrow x_{n+1} \leq \frac{1}{2}(2 + \frac{2}{1}) = 2$
Also by AM GM inequality, $x_n \geq \sqrt{2} \: \: \forall n$
Now all that I have to show is that $x_n  \geq x_{n+1} \: \forall n?$

At first one could show that $f(x)\geq 2$ for all $x>0$, where $f(x):=\frac12\left(x+\frac2{x}\right)$. This follows immediately from $f(x)^2-4= \frac14\left(x-\frac2{x}\right)^2\geq0$. But then $f(x)\leq x$ is equivalent to $\frac2{x}\leq x$ which is always  true when $x^2\geq 2$. But this holde true for $x_0=2$ by assumption, and thus by definition of $x_n$ for all $n$.
