If $A B=10$ and $B C=4$, then $(10-4) \leq A C \leq(10+4)$ by the triangle inequality. In the triangle inequality, the equality is only reached when the "triangle" $A B C$ is really a degenerate triangle, and $A B C$ are collinear. Simplifying, this means the smallest value $A C$ can be is 6 . Applying the triangle inequality on $A C D$ with $A C=6$ and $C D=3$, we know that $6-3 \leq A D \leq 6+3$ when $A C$ is minimized. If $A C$ were larger, then $A D$ could be larger, but we want the smallest $A D$ possible, and not the largest. Thus, $A D$ must be at least 3 , but cannot be smaller than 3 . Therefore, $B$ is the answer. 
Points $A$ and $B$ are 10 units apart. Points $B$ and $C$ are 4 units apart. Points $C$ and $D$ are 3 units apart. If $A$ and $D$ are as close as possible, then the number of units between them is ．(1996 AJHSME, Question 8) 

(a) $$0$$ 
(b) $$3$$ 
(c) $$9$$ 
(d) $$11$$ 
(e) $$17$$ 