Problem: Suppose $R>0$, $x_0 >0$ and $$x_{n+1}=\frac{1}{2}\left(\frac{R}{x_n}+x_n\right)$$ $n \geq 0$.
Prove,$$x_n - \sqrt{R} \leq \frac{(x_0-\sqrt{R})^2}{2^nx_0}$$ for $n \geq 1$.
My attempt: Since it looks like a neat closed form can't be found, it tried to use induction. The usual analysis of sequence, shows that sequence is decreasing and AM-GM reveals $x_n \geq \sqrt{R}$
Now, I assumed $$x_k\leq \frac{(x_0-\sqrt{R})^2}{2^kx_0}+\sqrt{R}$$
Combining with $\frac{1}{x_n} \leq \frac{1}{\sqrt{R}}$ i get,
$x_{k+1} < \frac{(x_0-\sqrt{R})^2}{x_02^{k+1}}+\sqrt{R}$ which is weaker than what is required.
Someone please help me. Any kind of proof would be appreciated.
Hint: Observe that $$x_{n+1} - \sqrt{R} = \frac{1}{2}\Big(\frac{R}{x_n} + x_n\Big) - \frac{2\sqrt{R}}{2} = \frac{1}{2}\Big(\frac{R + x_n^2 - 2\sqrt{R}{x_n}}{x_n} \Big) = \frac{1}{2}\frac{(x_n - \sqrt{R})^2}{x_n}$$