6. A cone and a cylinder, with their bottom faces on the same plane, have a common inscribed sphere. Let the volume of the cone be $V_{1}$, and the volume of the cylinder be $V_{2}$. Then the minimum value of $\frac{V_{1}}{V_{2}}$ is
6. $\frac{4}{3}$

Analysis: Let the radius of the common inscribed sphere be $R$, then the volume of the cylinder $V_{2}=2 \pi r^{3}$. Let the base angle of the cone be $2 \theta$, then the base radius of the cone is $r \cot \theta$, and the height is $r \cot \theta \cdot \tan 2 \theta$. Thus, $V_{1}=\frac{\pi a^{2} h}{3}=\frac{\pi r^{3} \cdot \cot ^{3} \theta \cdot \tan 2 \theta}{3}$, so $\frac{V_{1}}{V_{2}}=\frac{\cot ^{3} \theta \cdot \tan 2 \theta}{6}$. Let $t=\tan \theta, t \in(0,1)$, then it is known that $\frac{V_{1}}{V_{2}}=\frac{1}{3 t^{2}-3 t^{4}} \geq \frac{4}{3}$