The questions asks ...

Suppose that $\displaystyle R > 0$, $\displaystyle x_{0}>0$, and

$\displaystyle x_{n+1}= \frac{1}{2}(\frac{R}{x_{n}}+x_{n})$, $\displaystyle n\geq0$

Prove: For $\displaystyle n\geq1$, $\displaystyle x_{n}>x_{n+1}>\sqrt{R}$ and

$\displaystyle x_{n}-\sqrt{R}\leq\frac{1}{2^n}\frac{(x_{o}-\sqrt{R})^2}{x_{o}}$

MY attempt at a solution:

I really don't know how to go about this... at first i tried to show that if $\displaystyle x_{n}>x_{n+1}>\sqrt{R}$ is true then $\displaystyle x_{n}-x_{n+1} > 0$.

So that means

$\displaystyle x_{0}- \frac{1}{2}(\frac{R}{x_{0}}+x_{0})>0$

$\displaystyle \frac{x_{0}}{2}-\frac{R}{2x_{0}}>0$

$\displaystyle x_{0}-\frac{R}{x_{0}}>0$

My problem is that i don't know how big R in comparison to $\displaystyle x_{0}$, so unless i restrict R to be less than $\displaystyle x_{0}$ i can't prove that $\displaystyle x_{n}>x_{n+1}>\sqrt{R}$

Basically, i need help with this question... can anyone help ?

I'm doing a self study by myself so i would appreciate a detailed solution if possible.. please and thank u.