Prove $x_n \leq x_{n+1}$ for all $n$ by induction.
I am reading this example from "Understanding Analysis" by Abbott (page 10).
He says the multiple across the inequality by $1/2$ and then add 1 to get. $$\begin{equation} \frac{1}{2}x_n +1 \leq \frac{1}{2}x_{n+1} +1 \tag{1} \end{equation}$$ Which leads to $$x_{n+1} \leq x_{n+2} \tag{2}$$ I am not quite certain how he went from equation (1) to (2). I am also not clear why he multiplied across by $1/2$ and added 1; but the previous question might clear up the next question.
Shot of the book page:
The base case is clear ($x_1 = 1 \le 3/2 = x_2$). Now assume the induction hypothesis $x_n \le x_{n+1}$, and manipulate it as follows:
$$ x_n \le x_{n+1} \implies \frac{1}{2}x_n \le \frac{1}{2}x_{n+1} \implies \frac{1}{2}x_n +1 \le \frac{1}{2}x_{n+1} + 1 \implies x_{n+1} \le x_{n+2}. $$