The given equation is equivalent to $x_{n+1}x_{n+2} = \frac{x_nx_{n+1} +1}{2},$ so if we let $y_n = x_n x_{n+1}$ for $n \in \mathbb{N},$ then $(1-y_{n+1}) = \frac{1-y_n}{2}.$ Since $y_n$ is periodic (since $x_n$ is), this clearly implies that we must have $y_n \equiv 1,$ $\forall n \in \mathbb{N}.$ Therefore, this shows that our sequence is of the form $a, \frac{1}{a}, a, \frac{1}{a}, a, \cdots$ for some $a \in \mathbb{R}^{+},$ which clearly works. $\square$

XbenX wrote: Find all periodic sequences $x_1,x_2,\dots$ of strictly positive real numbers such that $\forall n \geq 1$ we have $$x_{n+2}=\frac{1}{2} \left( \frac{1}{x_{n+1}}+x_n \right)$$ https://latex.artofproblemsolving.com/6/d/4/6d4d4929393f83c641d6ec3466f0e599835333f5.png $2x_{n+2}x_{n+1}- x_{n+1}x_n=1$ . Let $x_{n+1}x_n=a_n$, so we find that $a_n=\frac{a_1-1}{2^{n-1}}+1$. If $x_{n+r}=x_n$ for some positive integer $r$, then $a_{r+1}=a_1$. $\implies$ $\frac{a_1-1}{2^{0}}+1=\frac{a_1-1}{2^{r}}+1$ $\implies$ $a_1=1$. It means $x_{n+1}x_n=1$ for all natural numbers $n$. $x_{n+1}x_n=x_{n+2}x_{n+1 }=1$ $\implies$ $x_{n+2}=x_n$ for all natural $n$. So $r=2$ and $x_1=a$, $x_2=\frac{1}{a}$ where $a$ is some constant. So we are done