Sorry for the annoyance but I have a proof to do which I don't find. I have $$ \begin{cases} u_0 = 0 \\ v_0 = 0 \end{cases} \\ \text{and} \begin{cases} u_{n+1} = \sqrt{3-v_n} \\ v_{n+1} = \sqrt{3+u_n} \end{cases} $$ I first had to show that these sequences are well defined which I did by showing first we have for all $n \in \mathbb N$: $$ \begin{cases} 0 \leq u_n \leq 3 \\ 0 \leq v_n \leq 3 \end{cases} $$ However, now I have to show this: $$ \begin{cases} |u_{n+1}-1| \leq |v_n-2| \\ |v_{n+1}-2| \leq \frac{1}{2}|u_n-1| \end{cases} $$ I thought about showing it by induction but I didn't manage to do it that way.
Thanks
Let us start with the first inequality. Notice that $$|u_{n+1}-1|\le |v_n-2|\\\Leftrightarrow |\sqrt{3-v_n}-1|\le|v_n-2|\\\Leftrightarrow|\sqrt{3-v_n}-1|\le |3-v_n-1|\\\Leftrightarrow |\sqrt{3-v_n}-1|\le|(\sqrt{3-v_n})^2-1|\\\Leftrightarrow |\sqrt{3-v_n}-1|\le |\sqrt{3-v_n}+1|\cdot |\sqrt{3-v_n}-1|\\\Leftrightarrow 1\le |\sqrt{3-v_n}+1|,$$ which is obviously true.
For the second, notice that $$2|v_{n+1}-2|\le |u_n-1|\\\Leftrightarrow 2|\sqrt{3+u_n}-2|\le |u_n-1|\\\Leftrightarrow 2|\sqrt{3+u_n}-2|\le|(\sqrt{3+u_n})^2-4|\\\Leftrightarrow 2|\sqrt{3+u_n}-2|\le |\sqrt{3+u_n}+2|\cdot |\sqrt{3+u_n}-2|\\\Leftrightarrow 2\le |\sqrt{3+u_n}+2|,$$ which is once again true.