Four balls are externally tangent to each other, and there is a smaller ball that is externally tangent to all four of these balls. Find the radius of this smaller ball.
As shown in the figure, construct a tetrahedron $ABCD$ with the centers of four spheres as vertices, where $AB=6$, $CD=4$, $AC=BC=AD=BD=5$. Let the midpoints of $AB$ and $CD$ be $F$ and $E$ respectively, and the center of the small sphere be $O$. By the symmetry of the figure, it is known that point $O$ lies on $EF$. Let the radius of the small sphere be $r$, then $OE=\sqrt{OC^2-EC^2}=\sqrt{(r+2)^2-2^2}=\sqrt{r(r+4)}, OF=\sqrt{OA^2-AF^2}=\sqrt{(r+3)^2-3^2}=\sqrt{r(r+6)}, EF=\sqrt{AE^2-AF^2}=\sqrt{AD^2-DE^2-AF^2}=\sqrt{5^2-2^2-3^2}=\sqrt{12}$. Since $EF=OE+OF$, we have $\sqrt{12}=\sqrt{r(r+4)}+\sqrt{r(r+6)}$. Rationalizing the right side gives $\frac{r}{\sqrt{3}}=\sqrt{r(r+6)}-\sqrt{r(r+4)}$. Adding the two equations yields $2\sqrt{r(r+6)}=\frac{r+6}{\sqrt{3}}$, solving for $r$ gives $r=\frac{6}{11}$.