From numerical test, I know $x=1$ is an attractive fixed point of the function
$$
f(x)=\frac12 \left(x+\frac{1}{x}\right),
$$
on $(0,\infty)$.  

Is there a way to prove it?

Since
$$
f'(x)=\frac12\left(1-\frac{1}{x^2}\right),
$$
then
$$
|f'(x)| < 1,
$$
on $[1,\infty)$, and Banach contraction principle gives the result.  

But how to proceed for $(0,1)$?  

Let $a\_0 \in (0,1)$ and choose 
$$
0< \epsilon < a\_0,
$$
then the derivative is bounded on $(\epsilon,\infty)$.  

But I don't see how to build a contraction on $(\epsilon,\infty)$, which gives the result.  

I tried
$$
g(x)=\frac{1}{M}f(x),
$$
with
$$
M=\sup\{x\in(\epsilon,\infty):|f'(x)|\}+1,
$$
but it doesn't work.