Problem: In $\triangle ABC$, $|BC|$ is longest side. $P$ is a point in the plane of $\triangle ABC$. Prove that $$ |PB|+|PC| \ge |PA|$$ and determine equality conditions.
Notes: I aimed to present this problem, the solution of which I know, to the benefit of the users, as it has an aesthetic aspect. I also shared the question to learn about different possible solutions myself. It is not made in a spirit like challenging to users. It is a post to contribute to accumulation that may benefit the community.
(For these reasons, I would appreciate if the problem is unblocked.)
According to Ptolemy's inequality, $$ PA \cdot BC \le PB \cdot AC + PC \cdot AB. $$ Taking into account $BC \ge AC$ and $BC \ge AB$, we get the desired inequality.
Equality holds if and only if $A,B,C,P$ are cocyclic (that's the equality condition for Ptolemy's inequality) and $BC = AC$, $BC = AB$. In other words, if the triangle $ABC$ is equilateral and $P$ lies on its circumcircle.