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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?

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    Related: http://math.stackexchange.com/questions/1716200/proving-a-sequence-converges. – 2017-01-08
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    What, wait, why $$\frac{x_n}{x_{n+1}}=\frac{x_n+\frac{1}{x_n}}{x_{n+1}+\frac{1}{x_{n+1}}}\ ?$$ – 2017-01-08
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    @Did See the edited question. – 2017-01-08

2 Answers 2

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You assume that "we are dealing with positive numbers only." However, this can be proven from the fact that $x_1 \geq 1 > 0$, so by induction, all $x_n > 0$ (since it is the product between a positive number, that is, $\frac 1 2$, and a sum of two positive numbers, that is, $x_n$ and $\frac{1}{x_{n-1}}$) and thus $a > 0$.

Other than forgetting to point this out, your reasoning is valid. However, this is not a proof. In a proof, you must start with a statement which is already known to be true, such as $x_n+\frac 1 {x_n} \geq 2$, and then work your way to what you are trying to prove, which is $1 \leq \frac{a}{\frac{1}{2}a+\frac{2}{a}}$. Therefore, in order to make this a proof, you need to go in the backwards order that you did originally.

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    I don't why this was downvoted. – 2017-02-02
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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 $$\frac{x_n}{x_{n+1}}=\frac{2x_n}{x_n+\frac{1}{x_n}}\leq 1$$

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    Would be nice if the downvoter(s) could explain their vote. – 2017-01-08