Four circles with radii 1, 2, 3, and $r$ are externally tangent to one another, where $r$ is the smallest radius.  Compute $r$.
Let $A,B,C,P$ be the centers of the circles with radii 1, 2, 3, and $r$, respectively.

[asy]
unitsize(1 cm);

pair A, B, C, P;

A = (0,0);
B = (0,3);
C = (4,0);
P = (20/23,21/23);

draw(Circle(A,1));
draw(Circle(B,2));
draw(Circle(C,3));
draw(Circle(P,6/23));
draw(A--B--C--cycle);
draw(A--P);
draw(B--P);
draw(C--P);

label("$P$", P, NE, UnFill);

dot("$A$", A, SW);
dot("$B$", B, NW);
dot("$C$", C, SE);
dot(P);
[/asy]

Then $ABC$ is a 3-4-5 right triangle.  By the Law of Cosines on triangle $PAB,$ \[
\cos\angle PAB=\frac{3^2+(1+r)^2-(2+r)^2}{2\cdot 3\cdot(1+r)} =
\frac{3-r}{3(1+r)}.
\]Similarly, \[
\cos\angle PAC= \frac{4^2+(1+r)^2-(3+r)^2}{2\cdot 4\cdot(1+r)} = \frac{2-r}{2(1+r)}.
\]Since $\angle PAB + \angle PAC = 90^\circ,$
\[\cos^2 \angle PAB  + \cos^2 \angle PAC = \cos^2 \angle PAB + \sin^2 \angle PAB = 1.\]Hence,
\[\left( \frac{3 - r}{3(1 + r)} \right)^2 + \left( \frac{2 - r}{2(1 + r)} \right)^2 = 1.\]This simplifies to $23r^2 + 132r - 36 = 0,$ which factors as $(23r-6)(r+6) = 0$.  Therefore, $r=\boxed{\frac{6}{23}}.$