sequences and series - Prove using induction

I have this math problem I'm kind of stuck on. Here's the question:

Define a sequences of real number with the definitions

$$\begin{align*} x_1 & = 3 \\ x_n &= \sqrt{2 x_{n-1}+1} \text{ for $n \ge 2$}. \end{align*}$$ Prove by induction that $x_n \ge x_{n+1}$.

I know that for induction you have to check the initial case. I set $G(n): x \ge x_{n+1}$. For my initial case I check $G(2)$. I get $\sqrt{7} > \sqrt{2\sqrt{7}+1}$, which is true. I know assume that $G(k)$ is true for some k $\in \mathbb{Z}^+, k \ge 2$. So I get my induction assumption $G(k): x_k \ge x_{k+1}$. Now I show that $G(k+1)$ is also true. So I have $G(k+1) = x_{k+1}\ge x_{k+2}$. I am not sure where to go from here. Thanks.

Answer

Simply use the definition of $x_{k+1}$.

We have to show

$x_{k+1}\geq x_{k+2}$.

By definition, this is equivalent to

$\sqrt{2x_k+1}\geq \sqrt{2x_{k+1}+1}$

Take the square this inequality: $2x_k+1\geq 2x_{k+1}+1$.

This is true by the induction assumption.