Threes and Fours - Futility Closet

A problem from the Tenth International Mathematical Olympiad, 1968:

Prove that in every tetrahedron there is a vertex such that the three edges meeting there have lengths which are the sides of a triangle.

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Well, three segments can make a triangle if and only if the longest of them is shorter than the other two combined. So consider a tetrahedron with vertices A, B, C, D, and let AB be the longest side. Now suppose there isn’t a vertex such that the three edges meeting there can form the sides of a triangle. The three edges that meet at vertex A are AB, AC, and AD. If those three edges won’t form a triangle, that means that AB ≥ AC + AD. Similarly, considering vertex B, we can conclude that BA ≥ BC + BD. Adding two those inequalities, we get 2AB ≥ AC + BC + AD + BD. But we also know that two of the tetrahedron’s faces are ABC and ABD, and each of these is a triangle, so AB < AC + BC and AB < AD + BD. Adding those two inequalities gives 2AB < AC + BC + AD + BD, which contradicts the inequality we just reached. So it can’t be the case that no vertex has edges that will form a triangle.