Proofs about a recursive sequence. Let $x_1=a>2$, and $$x_{n+1}=\frac{x_n^2}{2(x_{n}-1)}, n=1,2,...$$
Prove
(i) If $a\leq 3$, we have $x_n\leq 2+\frac1{2^{n-1}},(n=1,2...)$
(ii) If $a> 3$, then when $n\geq\frac{lg(\frac a3)}{lg \frac 43 }$, we have $x_{n+1}<3$.
My attempts:
I can use induction to show that $x_n>2$, and get $\frac{x_{n+1}}{x_n}=\frac{x_{n}}{2(x_n-1)}< 1$, where I checked the graph of the function $f(x)=\frac{x}{2(x-1)}$. For (i), I used $x_n>2$ and tried the induction to obtained that, under the assumption $x_n\leq 2+\frac1{2^{n-1}},(n=1,2...)$, we have $$x_{n+1}=\frac{x_n^2}{2(x_n-1)}\leq \frac{x_n^2}{2(2-1)}=\frac{x_n^2}{2}\leq \frac{(2+\frac1{2^{n-1}})^2}{2}.$$ My target next is to show $x_{n+1}\leq  2+\frac1{2^{n}}$ but I have no idea to go further. For(ii), I have no idea so far and any suggestions or hints would be very appreciated!
 A: The claimed inequality $$x_n \le 2 + \frac{1}{2^{n-1}}$$ suggests defining an auxiliary sequence $$y_n = x_n - 2,$$ hence the original recursion may be written
$$y_{n+1} = x_{n+1} - 2 = \frac{(x_n-2 + 2)^2}{2(x_n - 2 + 1)} - 2 = \frac{(y_n + 2)^2 - 4(y_n + 1)}{2(y_n + 1)} = \frac{y_n^2}{2(y_n + 1)}. \tag{1}$$  In turn, this suggests that when $y_n > 0$, $$y_{n+1} = \frac{y_n^2}{2(y_n+1)} \le \frac{y_n^2}{2y_n} = \frac{y_n}{2}. \tag{2}$$  Consequently, $$x_n - 2 = y_n  \le \frac{y_{n-1}}{2} \le \frac{y_{n-2}}{2^2} \le \cdots \le \frac{y_1}{2^{n-1}} = \frac{x_1 - 2}{2^{n-1}} = \frac{a-2}{2^{n-1}}.\tag{3}$$
Therefore, if $2 < a \le 3$, then $0 < a-2 \le 1$ and we have proven (i).
After illustrating the solution for the first part, I encourage you to reconsider how you might approach (ii).
A: For any $x>2,$ let $f(x)=\frac{x^2}{2(x-1)}.$ Then, $0<f(x)-2=\frac{(x-2)^2}{2(x-1)}<\frac{x-2}2$ hence by induction,
$$0<x_n-2<\frac{a-2}{2^{n-1}}.$$
This proves (i) but also (ii), since $2+2^n\ge3\left(\frac43\right)^n$ for every integer $n\ge1$ hence if $\left(\frac43\right)^n\ge\frac a3$ then $2+2^n\ge a$ i.e. $\frac{a-2}{2^n}\le1,$ which by the above implies $x_{n+1}<3.$
