This solution came from Ruth from Manchester High School for Girls. Well done Ruth!

Let the tetrahedron's vertices be $A$, $B$, $C$ and $D$ and the longest side be $AB$. If you assume that there is not a vertex where the three sides meeting at it could be the sides of a triangle, we must have $AC + AD < AB$ and $BC + BD < AB$ (otherwise the sides meeting at $A$ or $B$ could be the sides of a triangle). Therefore

$$AC + AD + BC + BD< 2 AB$$.Now since $ABC$ and $ABD$ are both triangles, we must have $AC + BC > AB$ and $AD + BD > AB$. Therefore

$$AC+ AD + BC + BD > 2 AB$$. This contradicts (1), so the initial assumption must be wrong. There is at least one vertex (one of $A$ or $B$) where the three sides meeting at it could be the sides of a triangle.