6. $(\sqrt{2}, \sqrt{2})$ Detailed Explanation: Let $P\left(x, \frac{1}{x}\right), A(a, b), B(-a,-b)$, then point $P$ is a moving point on the hyperbola $y=\frac{1}{x}$, and points $A$, $B$ are symmetric with respect to the origin. By the geometric properties of the hyperbola, the points $A, B$ that make $|P A|-|P B|$ a constant must be the foci of the hyperbola. It is known that the foci of the hyperbola $y=\frac{1}{x}$ are $(\sqrt{2}, \sqrt{2}),(-\sqrt{2},-\sqrt{2})$. Since $a, b>0$, we have $(a, b)=(\sqrt{2}, \sqrt{2})$.
6. For any $x \in(-\infty, 0) \cup(0,+\infty), \sqrt{(x-a)^{2}+\left(\frac{1}{x}-b\right)^{2}}$ $\sqrt{(x+a)^{2}+\left(\frac{1}{x}+b\right)^{2}}(a>0, b>0)$ is always a constant, then the pair $(a, b)=$ $\qquad$ .