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 .

(1)

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 .

(2)

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.