Prove that the sum of the lengths of the three altitudes of a triangle is less than the sum of the lengths of the three sides of the triangle.

1. Let $AP,$ $BQ,$ and $CR$ be the altitudes of $\triangle ABC.$
Thus, we have \begin{aligned} AP < AB && [\text { As AP is the perpendicular from the point A.}] \\ BQ < BC && [\text { As BQ is the perpendicular from the point B.}] \\ CR < AC && [\text { As CR is the perpendicular from the point C.}] \\ \end {aligned} Adding the three equations, we have \begin{aligned} & AP + BQ + CR < AB + BC + AC \\ \implies & \text{ Sum of the lengths of the altitudes } < \text{ Sum of the lengths of the sides } \end{aligned}