Proof explanation strict inequality with sequence

Question

Prove that if $2 \lt x_1 \lt 3$ and $x_{n+1} = 2 + \sqrt{x_n-2}$ for $n \in \mathbb{N}$ then $0 \lt x_n \lt x_{n+1}$

Assuming $x \in \mathbb{R}$

clearly $n = 1$

$x_1 \gt 0$ since $x_{n} \gt 2$

$x_{n+1} \gt 2$

So both $x_n$ and $x_{n+1} \gt 0 \space \space \space \forall n \in \mathbb{N}$

So now:

$x_n \lt 2 + \sqrt{x_n -2}$

$x_n - 2 \lt \sqrt{x_n - 2}$ $x = x_n - 2$

we have $x_n \lt \sqrt{x} \space \space \space \space \forall x \lt 1$

I don't know how to end the argument for some reason.

Answer 1

$2<x_n<3\implies 0<x_n-2<1 \implies 0<x_n-2<\sqrt{x_n-2}=x_{n+1}-2<1$

$\implies 2<x_n<x_{n+1}<3.$

Therefore $2<x_n<3\implies2<x_{n+1}<3$ and $x_n<x_{n+1}$.

Since $2<x_1<3,$ it follows that, for all $n\in\Bbb N,$ $2<x_{n+1}<3$ and $x_n<x_{n+1}.$

(Note that $0<x_n$ from the definitions of $x_1$ and $x_{n+1}.$)

Answer 2

You can show: $3 > x_n > 2, \forall n \ge 1$. Lets do this first. We have $3 > x_1 > 2$ ( given ). Assume $x_n > 2 \implies x_{n+1} = 2 + \sqrt{x_n - 2} > 2 + 0 = 2$, and $x_{n+1} < 2+\sqrt{3-2} = 3$ Thus $3 > x_n > 2, \forall n \ge 1$. This means $x_n > 0, \forall n \ge 1$. Next you show $x_n < x_{n+1}$. Indeed, $x_{n+1} - x_n = 2+\sqrt{x_n-2} - x_n= \sqrt{x_n-2}\left(1 - \sqrt{x_n-2}\right)> 0$ since $2 < x_n < 3\implies x_{n+1} > x_n$.