Prove that the sequence $x_{n+1}=\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)$ is not increasing Let $x_1\ge1$ and $x_{n+1}=\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)$. Prove that the sequence $\{x_n\}$ is not increasing, that is, $x_{n+2}\le x_{n+1}$.
My attempt was as follows: we have to prove that $1\le \frac{x_{n+1}}{x_{n+2}}$. Now:
$$\frac{x_{n+1}}{x_{n+2}}=\frac{x_n+\frac{1}{x_n}}{x_{n+1}+\frac{1}{x_{n+1}}}=\frac{x_n+\frac{1}{x_n}}{\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)+\frac{1}{\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)}}$$
let $a=x_n+\frac{1}{x_n}$ for the sake of convenience, then:
$$ \frac{x_n+\frac{1}{x_n}}{\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)+\frac{1}{\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)}}=\frac{a}{\frac{1}{2}a+\frac{2}{a}}$$
so for the claim that $1\le \frac{x_{n+1}}{x_{n+2}}$to hold, we must have:
$$ 1\le\frac{a}{\frac{1}{2}a+\frac{2}{a}}\rightarrow \frac{1}{2}a+\frac{2}{a}\le a\rightarrow \frac{1}{2}+\frac{2}{a^2}\le 1\rightarrow a^2\ge 4\rightarrow a\ge 2\rightarrow x_n+\frac{1}{x_n}\ge 2$$
which follows by AM-GM. 
My concern is, is this reasoning correct? I skipped some cases, assuming that we are dealing with positive numbers only - is my reasong rigorous? Or is there a serious flaw? 
Either way, maybe there is a cleaner and perhaps more clever way to prove the claim? 

The reasoning seems correct until the end when you write
$$x_n+\frac{1}{x_n}\geq 2$$ Now here you should prove that $$x_n+\frac{1}{x_n}\geq 2$$
Which is trivial by AM-GM.
Here's a similar approach note that by AM-GM
$$x_{n+1}=\frac{1}{2}(x_n+\frac{1}{x_n})\geq 1$$
hence we have that $x_n\geq 1$.
Now $$\frac{x_n}{x_{n+1}}=\frac{2x_n}{x_n+\frac{1}{x_n}}$$
But we have that $x_n\geq\frac{1}{x_n}$ because $x_n\geq 1$ that means $$\frac{x_n}{x_{n+1}}=\frac{2x_n}{x_n+\frac{1}{x_n}}\leq 1$$
