Tetra Inequalities

Try a proof by contradiction and use the Triangle Inequality which says that a triangle can be constructed with three 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 $AB$. 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 $B$?