Show the map $f(x)=\frac12 (x+1/x)$ has an attractive fixed point in $(0,\infty)$

Question

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.

Answer 1

Use the inequality $x+\frac{1}{x}\ge 2$, which can be proved many ways. For example, we can note that $\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2\ge 0$. Or else we can quote AM/GM.

Remark: The fixed point is in fact very attractive, since the derivative there is $0$.

Answer 2

Since $f(x)=f\left(\frac 1 x\right)$, once proven for $x > 1$ that $f^{(n)}(x)\xrightarrow[n\to\infty]{} 1$, then you get it for $x\in(0,1)$ "for free".