Inequality for each $a, b, c, d$ being each area of four faces of a tetrahedron We know 'triangle inequality'. I'm interested in the generalization of this inequality.
Here is my question.
Question: How can we represent a necessary and sufficient condition for each positive number $a, b, c, d$ being each area of four faces of a tetrahedron?
I've tried to get some kind of inequality, but I'm facing difficulty.
 A: The Triangle Inequality is an aspect of the Law of Cosines.
$$\begin{align}
a \le b + c \quad b \le c + a \quad c \le a + b \quad &\implies \qquad |b-c| \le a \le b+c\\
&\implies b^2 + c^2 - 2 b c \le a^2 \le b^2 + c^2 + 2 b c \\
&\implies \exists\;\theta, \; 0 \leq \theta \leq \pi \quad \text{s.t.} \quad a^2 = b^2 + c^2 - 2 b c \cos\theta 
\end{align}$$
where it turns out that "$\theta$" is exactly the angle that fits into the appropriate corner of the triangle. (Proof left to reader.)
For tetrahedra, we have this Law of Cosines involving face areas $W$, $X$, $Y$, $Z$ and dihedral angles $A$, $B$, $C$ (meeting at the vertex opposite face $W$) and $D$, $E$, $F$ (surrounding face $W$). For instance,
$$W^2 = X^2 + Y^2 + Z^2 - 2 Y Z \cos A - 2 Z X \cos B - 2 X Y \cos C$$
(with $A$ between faces $Y$ & $Z$, etc). Clearly, this gives the necessary condition
$$W \leq X + Y + Z$$
and its kin, although these are not sufficient.
Interestingly, when you've come to know tetrahedra like I know them, you realize that there are in fact seven faces to each of these things: the four familiar ("standard") ones, and three that I call "pseudo-faces". A pseudo-face is the projection of the tetrahedron into a plane parallel to a pair of opposite edges. I denote the areas of these $H$, $J$, $K$. 
More-interestingly, there's a Law of Cosines involving pseudo-faces:
$$\begin{align}
Y^2 + Z^2 - 2 Y Z \cos A \quad &= H^2 = \quad W^2 + X^2 - 2 W X \cos D \\
Z^2 + X^2 - 2 Z X \cos B \quad &= J^2 = \quad W^2 + Y^2 - 2 W Y \cos E \\
X^2 + Y^2 - 2 X Y \cos C \quad &= K^2 = \quad W^2 + Z^2 - 2 W Z \cos F 
\end{align}$$
which, together with the Law of Cosines above, proves this Sum-of-Squares identity:
$$W^2 + X^2 + Y^2 + Z^2 = H^2 + J^2 + K^2 \qquad(1)$$
Now, given seven ostensible areas (four standard and three pseudo), the Law of Cosines leads to Triangle-Inequality-like conditions, such as
$$|Y-Z| \leq H \leq Y+Z \qquad\qquad |W-X| \leq H \leq W + X \qquad (2)$$
(and likewise for $J$ and $K$). Of course, the areas must also satisfy the Sum-of-Squares identity $(1)$. But even this collection of conditions isn't sufficient to determine a tetrahedron. We need one more:
$$\begin{align}
0 \quad \leq \quad &2 W^2 X^2 Y^2 + 2 W^2 Y^2 Z^2 + 2 W^2 Z^2 X^2 + 2 X^2 Y^2 Z^2 + H^2 J^2 K^2 \\
&-H^2\left(W^2 X^2+Y^2 Z^2\right)
-J^2\left(W^2 Y^2+Z^2 X^2\right)
-K^2\left(W^2 Z^2+X^2 Y^2\right) \qquad (3)
\end{align}$$
When the right-hand side is in fact non-negative, it gives $81 V^4$, where $V$ is the volume of the tetrahedron.
Together, $(1)$, $(2)$, $(3)$ constitute my analogue of Menger's Theorem (which outlines conditions under which six edge-lengths can make a tetrahedron). For further information on this result, see my Bloog post "A Hedronometric Theorem of Menger".
FYI: The Bloog also has a number of other notes on "Hedronometry" ---my name for the dimensionally-enhanced trigonometry of tetrahedra--- both Euclidean and non-. (The earliest notes need some editing love. I was just using them for TeX practice waaaay-back-when. :)

So, one way to answer your question is this:

$W$, $X$, $Y$, $Z$ can be areas of faces of a tetrahedron if and only if there exist non-negative $H$, $J$, $K$ satisfying $(1)$, $(2)$, $(3)$.

A: $$a<b+c+d$$
$$b<c+d+a$$
$$c<d+a+b$$
$$d<a+b+c$$
Is that what you want?
This condition is necessary. You can't have sufficient condition with areas alone.
Let's investigate if a tetrahedron with triangle areas $a,b,c,d$ is possible if the conditions are true.
Let $a>b, a>c, a>d$. Make $a$ a regular triangle with side $A$. Now you must construct one more point on position $(x,y,z)$. Triangles $b,c,d$ must have heights $H_b,H_c,H_d$ such that $AH_b/2=b$ etc. Now we have three constraints and three variables, can you solve it?
A: Just an idea without writing any proof.
The volume $V$ of a tetrahedron $O-PQR$ can be represented by six edges $OP=p, OQ=q, OR=r, QR=l, RP=m, PQ=n$ as the following:
$$144V^2=p^2l^2(-p^2+q^2+r^2-l^2+m^2+n^2)+q^2m^2(p^2-q^2+rc^2+l^2-m^2+n^2)+r^2n^2(p^2+q^2-r^2+l^2+m^2-n^2)-l^2q^2r^2-p^2m^2r^2-p^2q^2n^2-l^2m^2n^2.$$
Let the right side of this expression be $F(p, q, r; l, m, n)$.
On the other hand, the necessary and sufficient condition for given four triangles $OPQ, ORQ, PRQ, OPR$ being the faces of a tetrahedron is that
$$F(p, q, r; l, m, n)>0.$$
By Heron's formula, 
$$4a=\sqrt{(p^2+q^2+n^2)^2-2(p^4+q^4+n^4)}, 4b=\sqrt{(q^2+r^2+l^2)^2-2(q^4+r^4+l^4)},$$
$$4c=\sqrt{(l^2+m^2+n^2)^2-2(l^4+m^4+n^4)}, 4d=\sqrt{(p^2+r^2+m^2)^2-2(p^4+r^4+m^4)}.$$
The problem is that it seems impossible to represent $F$ only by $a,b,c,d$.
