$a_1=\sqrt{6}$ , $a_{n+1} = \sqrt{6+a_n}$ A sequence $(a_n)$ is defined by $a_n > 0$ and $a_{n+1} = \sqrt{6+a_n}$ for $n\ge1$. Show that
$1)$ the sequence $(a_n)$ is monotone increasing if $0 < a_1 < 3$
$2)$ the sequence $(a_n)$ is monotone decreasing if $a_1 > 3$
I know how to verify it by substituting values for $n$ but to prove generally, I don't know.
I know that the limit of the sequence is $3$ which is obtained by the positive root of $x^2 - x -6 =0$.
We usually use $a_{n+1} - a_n$ to check if a sequence is monotone increasing our monotone decreasing. But here, since intervals are involved, I don't know how to proceed.
Thank you in advance.

If $a_n > 3$ then
$$a_n = \sqrt{a_n \cdot a_n} > \sqrt{3 a_n} = \sqrt{2 \cdot a_n + a_n} > \underbrace{\sqrt{6 + a_n}}_{= a_{n+1}} > \sqrt{6 + 3} = \sqrt{9} =3 $$
and if $0 < a_n < 3$ then
$$a_n = \sqrt{a_n \cdot a_n} < \sqrt{3 a_n} = \sqrt{2 \cdot a_n + a_n} < \underbrace{\sqrt{6 + a_n}}_{= a_{n+1}} < \sqrt{6 + 3} = \sqrt{9} =3 $$
by strict monotony of the square root (in each non-trivial step an $a_n$ is replaced by a $3$). The desired statement follows by complete induction on $n$.
