Two triangles $ABC$ and $A'B'C'$ are on the plane. It is known that each side length of triangle $ABC$ is not less than $a$, and each side length of triangle $A'B'C'$ is not less than $a'$. Prove that we can always choose two points in the two triangles respectively such that the distance between them is not less than $\sqrt{\dfrac{a^2+a'^2}{3}}$.
Problem
Source: 2022 Taiwan TST Round 1 Independent Study 1-G
Tags: Taiwan, geometric inequality, geometry, inequalities