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I'm stuck trying to understand part (ii) of the following question;

Fix a positive real number a > 0 and generate a sequence by setting $x_0 = 1$ and, $$x_n = \dfrac{1}{2}\bigg(x_{n-1}+\frac{a}{x_{n-1}}\bigg),$$ for $n = 1, 2, 3....$

(i) The arithmetic mean - geometric mean inequality says that $(\dfrac{x+y}{2}) \geq \sqrt{xy}$ for all $x, y \geq 0.$ Use this to show $x_n-x_{n+1}\geq0$ for all $n= 1, 2, 3,....$

Answer is just := $x_n=\dfrac{1}{2}(x_{n-1}+\dfrac{a}{x_{n-1}}) \geq \sqrt{x_{n-1}(\frac{a}{x_{n-1}})}=\sqrt{a}$

(ii) Use part (i) to show that $x_n - x_{n+1} \geq0$ for $n= 1, 2, 3...$

Solution = $x_n-x_{n+1} = x_n-\dfrac{1}{2}(x_n+\frac{a}{x_n})=\dfrac{(x_n)^2-a}{2x_n} \geq0.$

The solution tells me that $x_n -\dfrac{(x_n)^2-a}{2x_n} \geq0,$ but I don't understand why this has to be a positive number.

Thanks.

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    Because $x_n>\sqrt{a}>0 \Rightarrow x_n^2>a$. – 2017-01-04
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    In part (i) you should have “Use this to show $x_n\ge\sqrt{a}$”. – 2017-01-04
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    I will mention that you can find other posts about this sequence, see [Proof of Convergence: Babylonian Method $x_{n+1}=\frac{1}{2}(x_n + \frac{a}{x_n})$](http://math.stackexchange.com/q/82682) and other questions [linked there](http://math.stackexchange.com/questions/linked/82682). A [cobweb plot](http://math.stackexchange.com/questions/1884302/fixed-point-iterations-for-real-functions-depending-on-fx) might help you visualize the behavior of these iterations. – 2017-01-04

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In (i) you show that $x_n\ge\sqrt{a}$ for each $n\in\Bbb Z^+$, so you know that $x_n^2\ge a$ for each $n\in\Bbb Z^+$ and hence that $x_n^2-a\ge 0$ for each $n\in\Bbb Z^+$. That immediately implies that

$$\frac{x_n^2-a}{2x_n}\ge 0$$

for each $n\in\Bbb Z^+$, since you know that each $x_n$ is positive.

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    Hi, I understand the 2nd term is positive from your explanation, but how do I know that $x_n \geq \frac{(x_n)^2-a}{2x_n}$? Apologies if that information is in your answer, I just can't see it. I'm still new to this analysis stuff. – 2017-01-04
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    @Ryan: You don’t need to know that: what you need to know is that $x_n-x_{n+1}\ge 0$, and that’s true because $x_n-x_{n+1}=\frac{x_n^2-a}{2x_n}\ge 0$. – 2017-01-04
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From $x_n = \frac{1}{2}(x_{n-1}+\frac{a}{x_{n-1}}) $, we get $x_n^2 = \frac{1}{4}(x_{n-1}^2+2a+\frac{a^2}{x_{n-1}^2}) $, so that $x_n^2 -a = \frac{1}{4}(x_{n-1}^2-2a+\frac{a^2}{x_{n-1}^2}) = \frac{1}{4}(x_{n-1}-\frac{a}{x_{n-1}})^2 = \frac{(x_{n-1}^2-a)^2}{4x_{n-1}^2} $.

In particular, this shows that $x_n^2 > a$ and, if $y_n = x_n^2 -a$, then $y_n =\dfrac{y_{n-1}^2}{4(y_{n-1}+a)} $.

This shows that $y_n \le \frac14\min( y_{n-1}, y_{n-1}^2/a) $ which should help.

Note: None of this is original.