Tetrahedron is minimum energy conformation for atom with 4 electron domains in VSEPR theory Background

A student asked me to prove that the regular tetrahedron is the minimum energy geometry available to describe the locations of the electron domains in three dimensions – the prediction of valence shell electron pair repulsion (VSEPR) theory for a central atom with 4 mutually repelling electron domains. With the tetrahedron demonstrated, it could then be shown using vector multiplication that the angles between all substituent atoms is $109.5^{\circ}$.


Some ideas

I can think of a few ways to start this proof:


As mentioned above, we could define a regular tetrahedron and work backwards to see if the distance between all vertices is equal and maximized. Start with the vertices: A(0, 0, 0), B(1, 1, 0), C(1, 0, 1) and D(0, 1, 1), and a central point O($\frac{1}{2}$, $\frac{1}{2}$, $\frac{1}{2}$). With these, we can get vectors that point to each vertex. Call them $\vec{r_1}$ through $\vec{r_4}$. We remember that we can calculate the angle between any two vectors using the geometric and algebraic definitions of the dot product. I won't bore you with the details of this calculation but if anyone wishes to see it, I would be happy to share it. If you do it you will see that all the angles are equal and satisfy: $\cos^{-1}\left(-\frac{1}{3}\right) = \theta_i \approx 109.47122^{\circ}$. From here we would need to calculate the six distances bewteen the four vertices to show that the distances are all equal. 
Using multivariable calculus, we could treat this as an optimization problem subject to the constraint matrix, $g$; the sum square distance function, $f$; and use the optimization formalism described by Lagrange. This results in a system of 13 unknowns – the 4 points with three coordinates each plus $\lambda$ – and 13 equations – the four constraint equations plus the nine partial derivative equations from the Lagrange method: 


$$g = 1 = \left\{\begin{array}{c}g_1(x_1,y_1,z_1) = {x_1}^2 + {y_1}^2 + {z_1}^2\\
g_2(x_2,y_2,z_2) = {x_2}^2 + {y_2}^2 + {z_2}^2\\
g_3(x_3,y_3,z_3) = {x_3}^2 + {y_3}^2 + {z_3}^2\\
g_2(x_4,y_4,z_4) = {x_4}^2 + {y_4}^2 + {z_4}^2\\
\end{array}\right.$$
$$f = \mathop{\sum_{i=1}^{3}\sum_{j=1}^{3}}_{i&lt;j}\left[\left({x_i}-{x_j}\right)^2+\left({y_i}-{y_j}\right)^2+\left({z_i}-{x_j}\right)^2\right] $$
$$\nabla f=-\lambda \nabla g$$


An approach similar to the one outlined above could be used with spherical coordinates instead. This version reduces the set of variables from 13 to 8 (the four sets of two angles for each point $\theta$'s and $\phi$'s; note the spherical radius could be set to 1 or any other bond distance, $a$, so is not a variable). These could be further reduced using symmetry arguments. For example, if we take the first vertex at the point $(0, 0, 1)$, we know based on symmetry that if we look down the z axis at the xy-plane, the angle between the three other vectors must have an angle of $120^{\circ}$ with respect to one another. Let one of the other vertices have the point $(x, 0, -\sqrt{1 - x^2})$ – satisfying the radius of 1 unit constraint we have imposed. Now based on symmetry, we know that the other two points must be $(x\cos(\frac{2\pi}{3}),x\sin(\frac{2\pi}{3}),-\sqrt{1-x^2})$ and $(x\cos(\frac{4\pi}{3}),x\sin(\frac{4\pi}{3}),-\sqrt{1-x^2})$ using spherical symmetry. From here, we can write a formula for the total sum square distance between all the points and set the derivative with respect to $x$ to zero to find the maximum sum square distance. Notice we have reduced the problem with these symmetry arguments from a multivariable calculus problem to a calculus BC problem. Alternatively, we could use some other guess and check method to solve for $x$. Note, we should find the same answer using this method as the others above (and below).
Sketch of a Simpler Solution
There is one remaining way that is simpler than all of the above, and it has to do with circumscribing a tetrahedron with a cube.
I will leave it to the reader to work on this most beautiful version. The properties of the cube ensure that the distances are maximized, and some more geometry will show that the inscribed shape is indeed a regular tetrahedron. For a hint, check out this excellent answer to this question on the math stack exchange site.