Find all periodic sequences $x_1,x_2,\dots$ of strictly positive real numbers such that $\forall n \geq 1$ we have $$x_{n+2}=\frac{1}{2} \left( \frac{1}{x_{n+1}}+x_n \right)$$
1. We start with the given recurrence relation for the sequence \( \{x_n\} \):
   \[
   x_{n+2} = \frac{1}{2} \left( \frac{1}{x_{n+1}} + x_n \right)
   \]
2. To simplify the analysis, we multiply both sides of the equation by \( x_{n+1} \):
   \[
   x_{n+1} x_{n+2} = \frac{1}{2} \left( 1 + x_n x_{n+1} \right)
   \]
3. Let \( y_n = x_n x_{n+1} \). Then the equation becomes:
   \[
   y_{n+1} = \frac{1 + y_n}{2}
   \]
4. Since \( \{x_n\} \) is periodic, \( \{y_n\} \) must also be periodic. Let the period be \( T \). Then:
   \[
   y_{n+T} = y_n
   \]
5. For \( y_n \) to be periodic, it must satisfy the recurrence relation:
   \[
   y_{n+1} = \frac{1 + y_n}{2}
   \]
   This implies that \( y_n \) must reach a fixed point. Let \( y_n = y \) for all \( n \). Then:
   \[
   y = \frac{1 + y}{2}
   \]
6. Solving for \( y \):
   \[
   2y = 1 + y \implies y = 1
   \]
7. Therefore, \( y_n = 1 \) for all \( n \). This implies:
   \[
   x_n x_{n+1} = 1 \quad \forall n
   \]
8. Given \( x_n x_{n+1} = 1 \), we can write:
   \[
   x_{n+1} = \frac{1}{x_n}
   \]
9. This implies that the sequence \( \{x_n\} \) alternates between \( a \) and \( \frac{1}{a} \) for some \( a \in \mathbb{R}^+ \). Therefore, the sequence is of the form:
   \[
   a, \frac{1}{a}, a, \frac{1}{a}, \ldots
   \]
10. We verify that this form satisfies the original recurrence relation:
    \[
    x_{n+2} = \frac{1}{2} \left( \frac{1}{x_{n+1}} + x_n \right)
    \]
    For \( x_n = a \) and \( x_{n+1} = \frac{1}{a} \):
    \[
    x_{n+2} = \frac{1}{2} \left( \frac{1}{\frac{1}{a}} + a \right) = \frac{1}{2} (a + a) = a
    \]
    For \( x_n = \frac{1}{a} \) and \( x_{n+1} = a \):
    \[
    x_{n+2} = \frac{1}{2} \left( \frac{1}{a} + \frac{1}{a} \right) = \frac{1}{2} \left( \frac{1}{a} + \frac{1}{a} \right) = \frac{1}{a}
    \]
    Thus, the sequence \( a, \frac{1}{a}, a, \frac{1}{a}, \ldots \) satisfies the given recurrence relation.

The final answer is \( \boxed{ a, \frac{1}{a}, a, \frac{1}{a}, \ldots } \) for some \( a \in \mathbb{R}^+ \).