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 $B$?