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Let $a$ be a positive real number, and let $x_1 > \sqrt a$. Define the sequence $\{x_n\}$ where $x_{n+1} = \frac{1}{2}(x_n + \frac{a}{x_n})$

I would like to show that $x_n > \sqrt a$ for every $n$, but I am unable to show this. Can anyone give me any ideas?

2 Answers 2

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Hint: Try the arithmetic-geometric inequality and use induction to show that there cannot be equality at any $n$.

Answer in spoiler:

The base case holds by hypothesis. Now suppose that $x_n > \sqrt{a}$ for some $n$. Then $x_{n+1} = \frac{1}{2}\left(x_n + \frac{a}{x_n}\right) \ge \sqrt{x_n\frac{a}{x_n}} = \sqrt{a}$ where the inequality is an application of the arithmetic-geometric inequality. There is equality if and only if ${x_n} = \frac{a}{x_n}$ but by our induction hypothesis that is impossible since $x_n > \sqrt{a} \implies \frac{a}{x_n} < \sqrt{a} < x_n$ Therefore we must have $x_{n+1} > \sqrt{a}$ as required.

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    @Wayne1212 I think it's a matter of taste.2012-09-20
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It suffices to show that if $x>\sqrt{a}$, then $x+\frac{a}{x}>2\sqrt{a}$ (why?). Note that $\frac{d}{dx}(x+\frac{a}{x})=1-\frac{a}{x^2}$ is positive for $x>\sqrt{a}$, and that when $x=\sqrt{a}$ we have $x+\frac{a}{x}=2\sqrt{a}$. Thus $x+\frac{a}{x}>2\sqrt{a}$ for $x>\sqrt{a}$.