Here is the original problem:
A cube is cut into several parts all of them tetrahedra. What is the minimum possible number of tetrahedra obtained in this way?
Note that each face of the cube must contain at least two of the tetrahedras, let's say $T, T'$ with areas $A, A'$. Since $A+A'\leq a^2$ where $a$ is the length of the edge of the cube we know that $\text{Vol}(T)+\text{Vol}(T')\leq a^3/3$.
That is what I have done.
I think the answer is five. I have two questions. The first is how to finish this proof? here is the second:
I have an example for 5: if the cube is $ABCDA'B'C'D'$, then the five tetrahedras are $ACDD', ABCB', A'B'D'A, ACD'B', B'C'D'C$. Are there other examples? How many? (I mean really different examples, so not the reflections or transformation of my example.)