6. In the tetrahedron $P-ABC$, $PA=PB=a, PC=AB=BC=CA=b$, and $a<b$, then the range of $\frac{a}{b}$ is $\qquad$ .
6. $\left(\frac{\sqrt{6}-\sqrt{2}}{2}, 1\right)$.

Solution: Regardless of what positive number $b$ takes, $\triangle ABC$ always exists, and its height is $\frac{\sqrt{3}}{2} b$. For $\triangle PAB$ to exist, it must be that $b < PC$, i.e., $\sqrt{a^{2}-\frac{b^{2}}{4}}>b-\frac{\sqrt{3}}{2} b$.
Squaring both sides, we get $a^{2}>(2-\sqrt{3}) b^{2}$, hence $\frac{a}{b}>\sqrt{2-\sqrt{3}}=\frac{\sqrt{6}-\sqrt{2}}{2}$.
Therefore, the range of $\frac{a}{b}$ is $\left(\frac{\sqrt{6}-\sqrt{2}}{2}, 1\right)$.