$P$ is any point in the interior of triangle $ABC$. Prove that $AB + AC > BP + CP$.

1. Let us first mark the point $P$ in the interior of $\triangle ABC$.
2. Now, let us join line $BP$ and $CP$ and produce $CP$ to meet $AB$ at $Q$.
Thus in $\triangle ACQ,$ we have \begin{aligned} & AC + AQ > CQ \\ \implies & AC + AQ > CP + PQ && \ldots (1) \end{aligned} Similarly in $\triangle BPQ,$ we have \begin{aligned} & BQ + PQ > BP && \ldots (2) \end{aligned}
4. By adding (1) and (2), we get:\begin{aligned} & AC + AQ + BQ + PQ > CP + PQ + BP \\ \implies & AC + AQ + BQ > CP + BP && [\text{Subtrating PQ from both the sides}] \\ \implies & AC + AB > CP + BP && [\text{As AQ + BQ = AB}] \end{aligned}
5. Thus, we have $$\bf {AC + AB > CP + BP}$$.