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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.

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    $\begingroup$ Notice that $f(1/x) = f(x)$, so what happens to $x \in (0,1)$ after one iterations? $\endgroup$
    – Erick Wong
    May 16, 2013 at 2:12
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    $\begingroup$ Iterations of $f$ are what you get from applying Newton's method to finding the positive root of $g(x)=x^2-1$. Here is a link to a proof that Newton's method will always give a sequence converging to the root, for a class of functions that includes $g$: planetmath.org/newtonsmethodworksforconvexrealfunctions $\endgroup$ May 16, 2013 at 2:16
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    $\begingroup$ For this particular case, the iterate of $f$ can be expressed in a closed form: $$f^{\circ n}(x) = \underbrace{f\circ f \circ \cdots \circ f}_{n\text{ times}}(x) = \frac{1+\alpha^{2^n}}{1-\alpha^{2^n}} \quad\text{ where }\quad \alpha = \frac{x-1}{x+1}$$For any $x \in (0,\infty)$, $$|\alpha| < 1 \implies\lim_{n\to\infty} f^{\circ n}(x) = 1$$ $\endgroup$ May 16, 2013 at 3:59

2 Answers 2

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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 $0$.

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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".

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