# What is lower limit condition of a surface of a tetrahedron?

$S_1$, $S_2$, $S_3$, 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 $f(S_2,S_3,S_4)<S_1<S_2+S_3+S_4$

• – mathlove Jan 13 '16 at 12:09
• And achille hui showed there that given 4 positive numbers $S_1, S_2, S_3, S_4$, in order for them to be realizable as the four face areas of a non-degenerate tetrahedron, a necessary and sufficient condition is the four inequalities you write. – mathlove Jan 13 '16 at 12:30

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., $|b - c| < a$.
• Can we have only one lower condition ? or impossible. If it exists, it should be cyclic .for example, the lower condition could be $|S_2-S_3|+|S_3-S_4|+|S_2-S_4|<S_1$ but I do not know if it exists such cyclic condition or how to prove that it is impossible to eliminate 3 necessary condition into one condition. – Mathlover Jan 15 '16 at 6:51
• There's the maximum of the lower bounds, which is easily shown to be symmetric in $S_{2}$, $S_{3}$, $S_{4}$, but that seems unlikely to be aesthetically satisfactory.... – Andrew D. Hwang Jan 15 '16 at 15:04