<s>[INST] How to show that the following sequence is monotinacly increasing? given $$a_{1}=1, \quad a_{n+1}=\frac{2\left(2 a_{n}+1\right)}{a_{n}+3}$$ I am trying to show $a_{n}\leq a_{n+1}$
I am trying to show in induction, so my assumption is that $a_{n}\leq a_{n+1}$ and I want to show $a_{n+1}\leq a_{n+2}$ the only thing I could get is that $$a_{n+2}=\frac{2\left(2 a_{n+1}+1\right)}{a_{n+1}+3}>\frac{2\left(2 a_{n}+1\right)}{a_{n+1}+3}=a_{n+1}\cdot \frac{(a_n +3)}{a_{n+1}+3}$$
 [/INST] Notice that $a_n$ had better never be $-3$ or the recursion fails to be defined.  We wish to show \begin{align*}
a_n &\leq a_{n+1}  \\
    &= \frac{2(2a_n+1)}{a_n+3}  \\
    &= \frac{4a_n+2}{a_n+3}
\end{align*}
so
\begin{align*}
\begin{cases}
4a_n + 2 \geq a_n(a_n+3) ,& a_n+3 > 0 \\
4a_n + 2 \leq a_n(a_n+3) ,& a_n+3 < 0 
\end{cases}  \\
\begin{cases}
4a_n + 2 \geq a_n^2+3a_n ,& a_n > -3 \\
4a_n + 2 \leq a_n^2+3a_n ,& a_n < -3 
\end{cases}  \\
\begin{cases}
0 \geq a_n^2-a_n-2 ,& a_n > -3 \\
0 \leq a_n^2-a_n-2 ,& a_n < -3 
\end{cases}  \\
\begin{cases}
0 \geq (a_n - 2)(a_n + 1) ,& a_n > -3 \\
0 \leq (a_n - 2)(a_n + 1) ,& a_n < -3 
\end{cases}  \\
\end{align*}
Drawing sample points from $(-\infty,-3)$, $(-3,-1)$, $(-1,2)$, and $(2,\infty)$, we find that this is satisfied for $a_n < 3$ and $a_n \in (-1,2)$.
We have $a_1 = 1$.  To be monotonically increasing, we must have $a_n \in [1,2)$ for all $n \geq 1$.  (In particular, although we get at least one step of increasing if $a_n < -3$, if any $a_n$ after $a_1$ is less than $a_1 = 1$, the sequence can't be increasing.)  We have $\frac{\mathrm{d}}{\mathrm{d}x} \frac{2(2x+1)}{x+3} = \frac{10}{(x+3)^2} > 0$ (for $x \neq -3$), so is increasing on $[1,2)$.  This means $\frac{2(2x+1)}{x+3}$ is lower bounded at $x = 1$ and is upper bounded at $x = 2$.  We compute $\frac{2(2(1)+1)}{(1)+3} = \frac{3}{2}$ and $\frac{2(2(2)+1)}{(2)+3} = 2$, so $a_n \in [3/2,2)$ for all $n$.  Therefore, the sequence is monotonically increasing.
</s>