$a_1=\sqrt{6}$ , $a_{n+1} = \sqrt{6+a_n}$ 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.
 A: It is easy to show by induction that in case 1) $0<a_n<3$ for all $n$ and in case 2) $a_n>3$ for all $n$. Now
$$
a_{n + 1}  - a_n  = \sqrt {6 + a_n }  - a_n  = \frac{{6 + a_n  - a_n^2 }}{{\sqrt {6 + a_n }  + a_n }} = \frac{{(3 - a_n )(a_n  + 2)}}{{\sqrt {6 + a_n }  + a_n }}.
$$
Taking into account my first sentence, the monotonicity properties follow readily.
A: Let
$$f(x):=\sqrt{6+x}$$
The proof by @dxiv is perfect with his key expression that I write under the following form:
$$\dfrac{a_{n+1}−a_n}{a_n-a_{n-1}}=\dfrac{1}{\sqrt{6+a_n}+\sqrt{6+a_{n-1}}} \color{red}{\approx \dfrac{1}{2\sqrt{6+L}} = f'(L)}$$
with added terms in red, providing an approximation for large enough values of $n$. In this way, one (re)discovers the fact that the rate of convergence of an iterative sequence $a_{n+1}=f(a_n)$ is governed by the derivative of $f$ at its limit value, here $L=3$.
I would like here to give a complementary graphical "proof without words" (in fact I have added a lot of words!).

Fig. 1: The two curves with equations $y=f(x)$ and $y=x$ intersect in fixed point $(L,L)=(3,3)$ with two staircase patterns: ascending in red (with $a_0=0 < L$), descending in blue (with $a_0=6 > L$), both converging to fixed point.
where the upper image is a zoom on the lower image. Fixed point iteration
$$a_{n+1}=f(a_n)$$
is materialized by staircase patterns, each step joining points which are alternately on the curve and on the bissector line:
$$\underbrace{(a_n,f(a_n)=a_{n+1})}_{\text{on the curve}} \ \ \to \ \ \underbrace{(a_{n+1},a_{n+1})}_{\text{on the biss. line}} \ \ \to \ \  \underbrace{(a_{n+1},f(a_{n+1})=a_{n+2})}_{\text{on the curve}} \ \ \to \ \ \text{etc.}$$
If one starts below (resp. above) limit point $3$, we will have an ascending (resp. descending) staircase.
Remark: See this answer of mine to this question in a similar (but very particular) issue.
A: 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 $n$.
A: Just as a complement, this is a typical application of the fixed point theorem. One possible statement of the theorem is that if $g:[a,b]\to [a,b]$ is contractive, it has one and only one fixed point $z \in [a,b]$ and the iteration $x_{n+1} = g(x_n)$ converges to $z$, for any given $x_0\in [a,b]$. If $g$ is differentiable, the contractivity corresponds to the derivative being less than one in absolute value.
In this case, applying the theorem with $g(x)=\sqrt{6+x}$, for instance in the interval $[0,+\infty)$, you see  that the sequence converges for any $a_0 \ge 0$. Knowing that the sequence converges, it is easy to establish that $z = \sqrt{6+z}$, hence, $z = 3$. Furthermore, you can have error bounds like
$$
|a_n-3| \leq \frac{L^n}{1-L} |a_1-a_0|, \quad L = g'(0)=\frac{1}{2\sqrt{6}}.
$$
