In a book "Pi and the AGM" in 1987, authors, Jonathan Borwein and Peter Borwein, introduced a magical algorithm to compute $\pi$. However there is a problem that I couldn't understand and couldn't find any proof or explaination. It is related to rate of convergence of the algorithm. 


Borwein brothers suggested two sequences $\{x\_n\}\_{ n\ge 0}$ and $\{y\_n\}\_{ n\ge 1}$ like: 


\begin{equation}
x\_{0} := \sqrt{2} ~~~~,~~~~y\_{1}:= 2^{1/4} 
\end{equation}


\begin{equation}
x\_{n+1}=\frac{\sqrt{x\_n} ~+~ 1/\sqrt{x\_n} }{2} ~~~,~~~
y\_{n+1}=\frac{y\_{n} \sqrt{x\_n}~+~ 1/\sqrt{x\_n}}{y\_{n} +1}
\end{equation}


Then my question is about a proof for the next inequality:


\begin{equation} \label{eq}
 \forall n \in \mathbb{N}~,~~~\frac{x\_n -1}{y\_n -1} < 2- \sqrt{3}
\end{equation}


How can I prove this inequality?



Rate of Convergence of Borwein Algorithm for computing Pi
Even for different starting values around $1$, both $x\_n$ and $y\_n$ rapidly approach $1$. 


Define $\alpha\_n = x\_n-1$ and $\beta\_n = y\_n-1$. These will be very small quantities. $\beta\_5 \lt 10^{-40}$ and $\alpha\_5$ is smaller. We want to estimate $\alpha\_n/\beta\_n$.


$$\sqrt{x\_n} = \sqrt{1+\alpha\_n} = 1+\frac{1}{2}\alpha\_n - \frac{1}{8}\alpha\_n^2 + O(\alpha\_n^3).$$


$$1/\sqrt{x\_n} = 1/\sqrt{1+\alpha\_n} = 1-\frac{1}{2}\alpha\_n +\frac{3}{8} \alpha\_n^2 + O(\alpha\_n^3)$$


$$x\_{n+1} = \frac{1}{2}(\sqrt{x\_n} + 1/\sqrt{x\_n}) = 1+\frac{1}{8}\alpha\_n^2 +O(\alpha\_n^3) $$


So, $\alpha\_{n+1} = \frac{1}{8} \alpha\_n^2 + O(\alpha\_n^3)$.


$$\begin{eqnarray}y\_{n+1} &=& \frac{y\_n\sqrt{x\_n} + \sqrt{x\_n} - \sqrt{x\_n} + 1/\sqrt{x\_n}}{y\_n+1} \newline &=& \sqrt{x\_n} + \frac{-\sqrt{x\_n} +1/\sqrt{x\_n}}{y\_n+1} \newline &=&1+\frac{1}{2}\alpha\_n-\frac{1}{8}\alpha\_n^2 + O(\alpha\_n^3) + (-\alpha\_n+\frac{1}{2}\alpha\_n^2+O(\alpha\_n^3))(\frac{1}{2} - \frac{1}{4}\beta\_n + O(\beta\_n^2)) \newline &=& 1 + \frac{1}{8}\alpha\_n^2 + \frac{1}{4}\alpha\_n\beta\_n + O(\beta\_n^3)\end{eqnarray}$$


So, $\beta\_{n+1} = \frac{1}{8}\alpha\_n^2 + \frac{1}{4}\alpha\_n\beta\_n + O(\beta\_n^3)$.


Instead of $\alpha\_n/\beta\_n$, consider the reciprocal.


$\beta\_{n+1}/\alpha\_{n+1} = 1 + 2 \beta\_n/\alpha\_n + O(\alpha\_n).$ This ratio grows exponentially, and we only need to establish that it is greater than $1/(2-\sqrt{3}) = 2+\sqrt{3} \lt 4$. 


For example, $\beta\_4/\alpha\_4 = 99.53, \beta\_5/\alpha\_5 = 200.06$.


To turn this into a rigorous argument, you can use effective instead of asymptotic estimates for $\sqrt{1+\alpha\_n}$, $1/\sqrt{1+\alpha\_n}$, and $1/(2+\beta\_n)$ when $\alpha\_n$ and $\beta\_n$ are small. 


