jmc

If $AB=10$ and $BC=4$, then $(10-4)\le AC\le (10+4)$ by the triangle inequality. In the triangle inequality, the equality is only reached when the "triangle" $ABC$ is really a degenerate triangle, and $ABC$ are collinear. Simplifying, this means the smallest value $AC$ can be is $6$. Applying the triangle inequality on $ACD$ with $AC=6$ and $CD=3$, we know that $6-3\le AD\le 6+3$ when $AC$ is minimized. If $AC$ were larger, then $AD$ could be larger, but we want the smallest $AD$ possible, and not the largest. Thus, $AD$ must be at least $3$, but cannot be smaller than $3$. Therefore, $\boxed{3}$ is the answer. This answer comes when $ABCD$ are all on a line, with $A=0,B=10,C=6,$ and $D=3$.