Suppose the contrary that there is a tetrahedron $ ABCD$ that does not have any such vertex. So $ AB\ge AC+AD$ and $ AB\ge BC+BD$. Thus we have $ 2AB\ge AC+AD+BC+BD$. On the other hand, $ AB<AC+BC$ and $ AB<AD+BD$, which imply $ 2AB<AC+AD+BC+BD$. Contradiction. Q.E.D.