If $a$, $b$, $c$, $d$, $e$, and $f$ are the edge lengths of the tetrahedron, we can say WLOG that the labels of the vertices are $a + b + c$, $a + d + e$, $b + d + f$, and $c + e + f$. Now let us assume towards contradiction that the condition is not satisfied, or WLOG that $a > b + c$. Let the label of each vertex be $S$. Then $a > \frac{S}{2}$, so $b + c < \frac{S}{2}$ and $d + e < \frac{S}{2}$. In particular, at least one of $b + d$ and $c + e$, WLOG $b + d$, is less than $\frac{S}{2}$. But then $d = S - (b + d)$ is greater than $\frac{S}{2}$, so $c + e$ is also less than $\frac{S}{2}$. Then $b + d + c + e < S$ which is impossible since $b + d + c + e = S$, and we have a contradiction, as desired.