$S_1$, $S_2$, $S_3$, and $S_4$ are the areas of the four faces.
We know that a triangle has a condition for their edges $a$, $b$, $c$, so all edge length must satify
$$|a-b|<c<a+b$$ or $$|a-c|<b<a+c$$ or $$|b-c|<a<b+c$$
Is there any such limitations for a tetrahedron?
It is obvious that upper limit is
$$S_1<S_2+S_3+S_4$$
$$S_2<S_1+S_3+S_4$$
$$S_3<S_1+S_2+S_4$$
$$S_4<S_1+S_2+S_3$$
What is the lower limit condition for a tetrahedron as we have for a triangle $|a-b|<c$? How can the lower limit condition of a surface on a tetrahedron be defined by other surfaces such as $f(S_2,S_3,S_4)<S_1<S_2+S_3+S_4$
Thanks for answers
As you probably know, the reverse triangle inequality follows from $$ b < a + c,\qquad c < a + b $$ by rearranging to $b - c < a$ and $c - b < a$, i.e., $|b - c| < a$.
Analogously, the three necessary conditions \begin{align*} %S_{1} < S_{2} + S_{3} + S_{4}, \\ S_{2} &< S_{1} + S_{3} + S_{4}, \\ S_{3} &< S_{1} + S_{2} + S_{4}, \\ S_{4} &< S_{1} + S_{2} + S_{3} \end{align*} are equivalent to \begin{align*} %S_{1} < S_{2} + S_{3} + S_{4}, \\ S_{2} - (S_{3} + S_{4}) &< S_{1}, \\ S_{3} - (S_{2} + S_{4}) &< S_{1}, \\ S_{4} - (S_{2} + S_{3}) &< S_{1}. \end{align*} Combining the first two, \begin{align*} |S_{2} - S_{3}| - S_{4} &< S_{1}, \\ S_{4} - (S_{2} + S_{3}) &< S_{1}. \end{align*} There are six other necessary inequalities obtained by cyclic permutation of all four indices.
Is the volume of a tetrahedron determined by the surface areas of the faces? linked by mathlove in the comments addresses sufficient conditions.