Dlanod is aiming higher

We want to find a Steiner minimal tree for the vertices of a regular tetrahedron of side length $1.8$ km.

My intuition would be to create a Steiner minimal tree between two vertices and the centroid of the tetrahedron, and then reflect this tree through the centroid and rotate it through $90^\circ$ to create a Steiner minimal tree between the other two vertices and the centroid.

This system can be shown to have length $1.8\big(\sqrt{3} + \tfrac{1}{\sqrt{2}}\big) = 4.39048366$ km, making Dlanod's task possible.

Without the ambition of finding the Steiner minimal tree, we merely attempt to meet the budget constraint, using the 2D-case as a guide. Working with a distance of 2 km rather than 1.8 km at first, for convenience, we place our four hotels at $(\pm 1,0,0)$ (the "lower hotels") and $(0,\pm 1, \sqrt{2})$ ("the upper hotels") . Our intuition and our understanding of the 2D-case (see Figure 7 here ) suggest to introduce two symmetric junction points $(0,0,z)$ and $(0,0, \sqrt{2}-z)$ , connect $(0,0,z)$ to the two "lower" hotels, and connect $(0,0, \sqrt{2}-z)$ to the two "upper" hotels. The total length of the five tubes in this design turns out to be $f(z)=4\sqrt{1+z^2}+\sqrt{2}-2z$ . It is a straightforward calculus exercise to show that the minimum of $f(z)$ is $\sqrt{2}+2\sqrt{3}$ , attained when $z=\frac{1}{\sqrt{3}}$ . If the distance between the hotels is required to be 1.8 km rather than 2 km, then the corresponding network has a length of $0.9( \sqrt{2}+2\sqrt{3})\approx 4.39$ km, meeting the budget constraint. Thus, $\boxed{Yes}$ , the task can be accomplished.

The attached graph is shifted down by $\frac{1}{\sqrt{2}}$ , for symmetry.