Lemma Given 4 positive numbers $A_1, A_2, A_3, A_4$, in order for them to be
realizable as the four face areas of a non-degenerate tetrahedron, a necessary and
sufficient condition is the existence of 4 unit vectors $\hat{n}_1, \hat{n}_2, \hat{n}_3, \hat{n}_4$
( no 2 lies on the same line, no 3 lies on the same plane) such that
$$A_1 \hat{n}_1 + A_2 \hat{n}_2 + A_3 \hat{n}_3 + A_4 \hat{n}_4 = \vec{0}$$
The necessary part is standard vector analysis. One can use the unit normal vectors for the faces as $\hat{n}_i$. I will skip this part of proof.
For the sufficient part, let's say we indeed have 4 such unit vectors.
Now define 4 planes $P_i, i = 1,\ldots, 4$ by
$$P_i = \Big\{\; \vec{x} \in \mathbb{R}^3 : \vec{x} \cdot \vec{n}_i = \frac{1}{A_i}\;\Big\}$$
and consider the intersections of any 3 of the planes:
$$\vec{x}_1 = P_2 \cap P_3 \cap P_4,\;\;
\vec{x}_2 = P_3 \cap P_4 \cap P_1,\;\;
\vec{x}_3 = P_4 \cap P_1 \cap P_2,\;\;
\vec{x}_4 = P_1 \cap P_2 \cap P_3$$
It is clear $\vec{x}_2, \vec{x}_3, \vec{x}_4$ lies on the plane $P_1$ and hence satisfy $\vec{x}_i \cdot \hat{n}_1 = \frac{1}{A_1}$ for $i \ne 1$. In general, we have
$\vec{x}_i \cdot \hat{n}_j = \frac{1}{A_j}$ whenever $i \ne j$. Notice
$$\vec{x_1} \cdot \hat{n}_1 = -\frac{1}{A_1} \vec{x_1} \cdot ( A_2 \hat{n}_2 + A_3 \hat{n}_3 + A_4 \hat{n}_4 ) = -\frac{1}{A_1} \left( \frac{A_2}{A_2} + \frac{A_3}{A_3} + \frac{A_4}{A_4} \right) = -\frac{3}{A_1}$$
We can conclude the distance of $\vec{x}_1$ from the plane $P_1$ is $\frac{4}{A_1}$. One way to interpret this result is if we from a tetrahedron $X$ from $\vec{x}_1, \vec{x}_2, \vec{x}_3$ and $\vec{x}_4$. The height of the tetrahedron with respect to the
face opposite to $\vec{x}_1$ (i.e. the one contained in $P_1$ ) is $\frac{4}{A_1}$. Similar conclusions can be drawn for other vertices
$\vec{x}_2, \vec{x}_3$ and $\vec{x}_4$.
Recall the four face areas of a tetrahedron is inversely proportional to the corresponding heights, we find the four face areas of $X$ has the right ratios:
$$ \text{Area}(X\cap P_1) :
\text{Area}(X\cap P_2) :
\text{Area}(X\cap P_3) :
\text{Area}(X\cap P_4) = A_1 : A_2 : A_3 : A_4$$
By scaling $X$ with right amount, we can realize $A_1, A_2, A_3, A_4$ as the four face areas of a tetrahedron.
A trivial corollary of the lemma is if $A_i$ are realizable face areas, then
we have inequalities like:
$$A_1 = |A_1 \hat{n}_1 | = |A_2 \hat{n}_2 + A_3 \hat{n}_3 + A_4 \hat{n}_4| < A_2 + A_3 + A_4$$
The inequality is strict because no two $\hat{n}_i$ is lying on the same line.
If we group these inequalities together, we find
Another set of necessary condition for $A_i$ to be realizable as face areas:
$$\begin{cases} A_1 < A_2 + A_3 + A_4\\ A_2 < A_3 + A_4 + A_1\\ A_3 < A_4 + A_1 + A_2\\ A_4 < A_1 + A_2 + A_3\tag{*1} \end{cases}$$
It turns out $(*1)$ is also a sufficient condition.
From $(*1)$, we can deduce two inequalities:
$$(A_1 + A_2)^2 - (A_3 - A_4)^2 > 0\quad\text{ and }\quad(A_1 - A_2)^2 - (A_3 + A_4)^2 < 0$$
For $\theta \in [0,\pi]$, if we define a function $f(\theta)$ by:
$$f(\theta) = (A_1^2 + 2A_1A_2\cos\theta + A_2^2) - (A_3^2 - 2A_3A_4\cos\theta + A_4^2)$$
The above two inequalities implies $f(0) > 0$ and $f(\pi) < 0$. This means we can
find a $\theta \in (0,\pi)$ such that $f(\theta) = 0$. Now start with any two unit vectors
$\hat{n}_1$ and $\hat{n}_3$. If one rotate $\hat{n}_1$ for an angle $\theta$ to get a new unit vector $\hat{n}_2$ and rotate $\hat{n}_3$ for an angle $\pi - \theta$ to get another
new vector $\hat{n}_4$, we will have:
$$\begin{align}|A_1 \hat{n}_1 + A_2 \hat{n}_2|^2 = &
(A_1^2 + 2A_1A_2\cos\theta + A_2^2) =
(A_3^2 - 2A_3A_4\cos\theta + A_4^2)\\ = &
|A_3 \hat{n}_3 + A_4 \hat{n}_4|^2\end{align}$$
It is then clear a further rotation of $\hat{n}_3$ and $\hat{n}_4$ will allow us to produce
4 unit vectors satisfying the condition in Lemma and hence $A_1, A_2, A_3, A_4$ are realizable as face areas of a tetrahedron.