Does there exist a tetrahedron, so that every edge is the side of an obtuse angle of a face? I have the following the question with me:

"Does there exist a tetrahedron, so that every edge is the side of an obtuse angle of a face?"

The problem looks easy but I am unable to prove it. Any help please
 A: Let the tetrahedron $ABCD$ be such that edge $\overline{BC}$ is opposite an obtuse angle at $A$ in face $\triangle ABC$. If the other edges of that face —$\overline{AB}$ and $\overline{AC}$— are both opposite obtuse angles, then those angles must be at $D$ in faces $\triangle ADB$ and $\triangle ADC$. But then, $\overline{AD}$, the edge common to those faces, cannot be opposite an obtuse angle.
A: The tetrahedron has only four faces, so at most four obtuse angles. At most four of the six edges can be opposite to these angles. 
A: 
Let $AC$ and $BC$ form an obtuse angle in $\triangle ABC$. Then $AB$ has to be a side of the obtuse angle of $\triangle ABD$. Assume that $\angle DAB$ is obtuse. Then $BD$ is a side of the obtuse angle of $\triangle BCD$. If $\angle BDC$ is obtuse, then $BC > BD > AB$, a contradiction.
Therefore $\angle CBD$ is obtuse and $CD$ forms an obtuse angle either with $AC$ or with $AD$. In the first case, $AD > CD > BD$, a contradiction. In the second case, $AC > CD > BD > AB$, also a contradiction.
A: Whenever $AB$ is the side of an obtuse angle in $\triangle ABC$, one of the other sides is longer ($BC$ if the obtuse angle is $\angle A$, or $AC$ if the obtuse angle is $\angle B$). 
So the longest edge of the tetrahedron can't be the side of an obtuse angle.
