<s>[INST] how to show that $a_{n+1}=\sqrt{6+a_{n}}$ is bounded? I have to show that the sequrnce $\{a_n\}$ with $a_{n+1}=\sqrt{6+a_{n}}$ is bounded, i know : $a_{n-1}<a_{n}$ because i demonstrated this. I tied to use this relation: $ab\leq \frac{a^2}{2}+\frac{b^2}{2}$. I don't kow how to increase $\sqrt{6+a_{n}}$
 [/INST] Claim. If $a_n$ is increasing, then $a_n<3$.
Otherwise, if $a_n\ge 3$, for some $n\in\mathbb N$, then
Case A. $a_n=3$, then $a_{n+1}=\sqrt{a_n+6}=\sqrt{3+6}=\sqrt{9}=3$. Hence $a_n$ is not increasing.
Case B. $a_n>3$. Then $a_{n+1}>a_n>3$ and
$$
a_{n+1}-a_n=\sqrt{a_n+6}-a_n=\frac{\sqrt{(a_n+6}-a_n)(\sqrt{a_n+6}-a_n)}{\sqrt{a_n+6}+a_n}\\=\frac{a_n+6 -a_n^2}{\sqrt{a_n+6}+a_n}=-\frac{(a_n+2)(a_n-3)}{\sqrt{a_n+6}+a_n}<0.
$$
Contradiction.
Hence, $a_n$ is upper bounded by $3$ and lower bounded by $a_1$.
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