For some reason, this stumped me for sometime before I found this out.
Problem: Let $D$ be some point inside $\triangle ABC$. Prove that $BA+CA>BD+CD$.
Extend the line segment $BD$ to meet $AC$ at $E$. We have $AB+AE>BE$ by $\triangle$ inequality.
Adding $EC$ on both sides,
again applying $\triangle$ inequality ($DE+EC>DC$)