# prove by induction the following statement [duplicate]

I want to test the following using mathematical induction.

If $$x_1=1$$, $$x_2=2$$ and $$x_{n+2}=\frac{1}{2}(x_n+x_{n+1})$$ prove that $$1 \leq x_n \leq 2$$ for all $$n$$ natural

My idea was to do induction on n, for this I considered the following

basic step

Let $$n=1$$, then $$x_{1+2}=x_3=\frac{1}{2}(x_1+x_2)=\frac{1}{2}(2+1)=\frac{1}{2}(3)=\frac{3}{2}$$, and clearly the equality is verified.

Now, for the inductive step, we can consider as hypothesis of induction that $$1 \leq x_n \leq 2$$ with $$x_{n+2}=\frac{1}{2}(x_n+x_{n+1})$$, $$x_1=1$$ and $$x_2=2$$

Thus, let $$x_{(n+1)+2}=x_{n+3}=\frac{1}{2}(x_{n+1}+x_{n+2})=\frac{1}{2}(x_{n+1}+\frac{1}{2}( x_n + x_{n+1}))$$

Now, it is not very clear to me how to accommodate the inequality to conclude the demonstration, any suggestions?

Use strong induction to establish that $$x_n$$ and $$x_{n+1}$$ are in $$[1,2]$$ and then just bound $$x_{n+2} = \frac{x_n+x_{n+1}}{2},$$ which is their average...