Prob. 17, Chap. 3 in Baby Rudin: For $\alpha > 1$, how to obtain these inequalities from this recurrence relation? Here's Prob. 17, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition: 

Fix $\alpha > 1$. Take $x_1 > \sqrt{\alpha}$, and define $$x_{n+1} = \frac{\alpha + x_n}{1+x_n} = x_n + \frac{\alpha - x_n^2}{1+x_n}.$$ 
(a) Prove that $x_1 > x_3 > x_5 > \cdots$. 
(b) Prove that $x_2 < x_4 < x_6 < \cdots$. 
(c) Prove that $\lim x_n = \sqrt{\alpha}$. 

My effort: 
From the recursion formula, we can obtain 
$$
\begin{align}
x_{n+1} &= \frac{ \alpha + x_n}{1+ x_n} \\
&= \frac{ \alpha + \frac{\alpha + x_{n-1}}{1+x_{n-1}} }{ 1 + \frac{\alpha + x_{n-1}}{1+x_{n-1}} } \\
&= \frac{ (\alpha + 1) x_{n-1} + 2 \alpha   }{ 2x_{n-1} + ( 1 + \alpha ) } \\
&= \frac{\alpha+1}{2} + \frac{2 \alpha - \frac{(\alpha+1)^2}{2} }{2x_{n-1} + ( 1 + \alpha ) } \\ 
&= \frac{\alpha+1}{2} + \frac{ \alpha - \frac{\alpha^2+1 }{2} }{2x_{n-1} + ( 1 + \alpha ) }.  
\end{align}
$$
What next? 

$$\text {Let } a_n=(\sqrt \alpha)\tan (d_n+\pi /4) \text { with } d_n\in (0,\pi /4).$$ Using $\tan (x +\pi /4)=(1+\tan x)/(1-\tan x)$ when $\tan x \ne 1,$ we arrive, after some calculation, that $$\tan d_{n+1}=k \tan d_n$$  $$ \text { where } k=\frac {1-\sqrt {\alpha} }{1+\sqrt {\alpha}}.$$ Observe that $0>k>-1$ because $\alpha >1.$
Remark. I tried this substitution by heuristic analogy with a substitution for Heron's method: If $\sqrt {\alpha} \ne b_1>0$ and $b_{n+1}=(b_n+\alpha /b_n)/2,$  then $b_n>\sqrt {\alpha}$ for $n\geq 2.$ Let $b_n=\sqrt {\alpha} \coth c_n$ for $n\geq 2$, with  $c_2>0.$ Then $c_n=2^{n-2}c_2$ for $n\geq 2.$ In this Q I tried $\tan$ because $\tan d_n$ alternates $\pm. $
