This problem turns out to be a Tough Nut though it is not hard! You are asked to prove that in every tetrahedron there is a vertex such that the three edges meeting there have lengths which could be the sides of a triangle.

Try a proof by contradiction and use the triangle inequality which says that a triangle can be constructed with 3 given segments for sides if and only if the sum of the lengths of any two exceeds the length of the third. (For example the lengths 2, 3 and 7 cannot make the sides of a triangle because $2+3<7$). One more hint, one of the edges of the tetrahedron must be the longest and, without loss of generality, you can label this edge $\mathrm{AB}$. Now, if you are using a proof by contradiction, what can you say about the 3 edges meeting at $A$ and similarly about the three edges meeting at $B$?