Constructing triangles using sides of tetrahedron I'm trying to solve the following:

Using six sides of tetrahedron, construct two triangles.

My try:
Let $SACB$ be a tetrahedron and denote sides as shown in image. Assume that $SC$ is longest side of the tetrahedron, i.e if we denote $SC=d$, then $d\ge w,x,y,z$.
$\hspace{6cm}$
Using the triangle inequality we have \begin{align*}x+y&\ge d\tag{1} \\ z+w&\ge d \tag{2}\end{align*}
Sum $(1)$ and $(2)$ - $\displaystyle \frac{x+y}{2}+\frac{z+w}{2}\ge d$.
It is easy for me to show that $\displaystyle d+\frac{z+w}{2}\ge \frac{x+y}{2}$ and $\displaystyle d+\frac{x+y}{2}\ge \frac{z+w}{2}$, hence we can construct a triangle with sides $\displaystyle d,\frac{x+y}{2},\frac{z+w}{2}$.
Now, denote $AB=\ell$. As it seems to me I need to show that we can construct another triangle with sides $\ell$ and some combination of $\displaystyle \frac{x}{2},\frac{y}{2},\frac{z}{2},\frac{w}{2}$. I couldn't manage to show it.
Any hint or a feedback will be appreciated, thank you!
 A: A notational preamble ...
We know that triangle with (non-negative) edge-lengths $x$, $y$, $z$ exists if and only if the tripartite Triangle Inequality holds:
$$x \leq y + z \qquad y \leq z + x \qquad z \leq x + y $$
One can show that the condition is equivalent to
$$\operatorname{heron}(x, y, z) \geq 0$$
where
$$\begin{align}
\operatorname{heron}(x,y,z) &:= (x+y+z)(-x+y+z)(x-y+z)(x+y-z) \\
&\,= -x^4 - y^4 - z^4 + 2 x^2 y^2 + 2 y^2 z^2 + 2 z^2 x^2 
\end{align}$$
Note that the heron is strictly positive for a non-degenerate triangle, but I choose to be inclusive.
(The product is called the "heron" because it's the heart of Heron's Formula for the area of the corresponding triangle (if it exists): $\frac{1}{4}\sqrt{\operatorname{heron}(x,y,z)}$, which highlights the necessity (though not the sufficiency) for the product to be non-negative. The $(x+y+z)$ factor isn't strictly necessary for testing triangle viability, but it streamlines the expanded form, which helps with an implicit calculation below.)

Now to the problem at hand ...
If we could say that, for every vertex of a tetrahedron, the edges meeting that vertex form a triangle, we'd be done: use the edges about any vertex to form one triangle, and use the edges about the opposite face to form the second triangle (which they clearly already do!). Unfortunately, we can't say that. (There are lots of tetrahedra such that the edges meeting at a vertex are, say, $1$, $1$, $1000$. No triangle there.) However, we can say that, for at least one vertex of any given tetrahedron, the edges meeting that vertex form a triangle. Here's why ...
Let our tetrahedron have edges $a$, $b$, $c$ about a vertex, with respective opposite edges $d$, $e$, $f$. One readily verifies that

The sum of "face herons'' equals the sum of "vertex herons".

which is to say
$$\begin{align}
&\quad\; \operatorname{heron}(d,e,f) + \operatorname{heron}(d,b,c) + \operatorname{heron}(a,e,c) + \operatorname{heron}(a,b,f) \\
&= \operatorname{heron}(a,b,c) + \operatorname{heron}(a,e,f) +
\operatorname{heron}(d,b,f) + \operatorname{heron}(d,e,c)
\end{align} \tag{$\star$}$$
(You don't even have to multiply everything out yourself. Consulting the expanded formula given above, it's obvious that both sides of $(\star)$ contain two copies of the negative of the fourth power of each length; only-slightly-less-obviously, the various cross terms match: for instance, $a^2$ gets paired with $b^2$, $c^2$, $e^2$, $f^2$ on each side.)
Since "face herons" —calculated with edge-lengths from legitimate triangles— are individually non-negative, so is their sum; relation $(\star)$ then implies that at least one "vertex heron" must be non-negative: the edges meeting at the corresponding vertex must form a triangle. (More-specifically: if the faces are non-degenerate triangles, the edges at some vertex must form a non-degenerate triangle.) $\square$

We have shown that it's always possible to partition a tetrahedron's edges to form two triangles. If the task is to provide an explicit partition, then there's more work to do. I'll leave that as an exercise to the reader.
