6. $a, b$ are positive real numbers, $x_{0}=a, x_{1}=b, x_{n+1}=$ $\frac{1}{2}\left(x_{n-1}+\frac{1}{x_{n}}\right)$, where $x \in N$. If this sequence is a periodic sequence, then $a, b$ must satisfy the condition
$6 . a b=1$.
$x_{n} x_{n+1}=\frac{1}{2}\left(x_{n-1} x_{n}+1\right)$, let $y_{n}=x_{n} x_{n+1}$, $y_{n}=\frac{1}{2}\left(y_{n-1}+1\right)$. If $x_{n}$ is periodic, then $y_{n}$ is also periodic. If $y_{n-1}>1$, then $y_{n-1}>y_{n}>1$. Therefore, for $y_{n}$ to be periodic, it must be that $y_{n}=1$ for any non-negative integer $n$. Thus, $a b=y_{0}=1$.