The original problem posed was the following:
On a sphere of radius $1$ are given four points $A,B,C,D$ such that
$$AB\cdot AC\cdot AD \cdot BC \cdot BD \cdot CD = \frac{2^9}{3^3}$$
I am aware of a proof to this that uses the Lagrange identity. I am more interested, however, if it is possible to derive the same result using Jensen's inequality.
My idea:
If we could show the latter half of the below inequality we would be done
$$(AB\cdot AC\cdot AD \cdot BC \cdot BD \cdot CD)^{1/6} \le 1/6 (AB+AC+AD+BC+BD+CD) \le \frac{2^{2/3}}{3^{1/2}}$$
So I wanted to show that $AB+AC+AD+BC+BD+CD \le \frac{6\cdot2^{2/3}}{3^{1/2}}$, and furthermore that equality could only hold if the tetrahedron was regular.
This proved too difficult at first so I reduced the dimensions of the problem: i.e. I attempted to maximize the perimeter of a triangle circumscribed in the unit circle.
The 2-d case:
Firstly, it is clear that a triangle that maximizes the perimeter must contain in its interior the center of the circle. Next, we see that the angles from the center of the circle to each of the chords formed by the legs of the triangle (call these angles $x,y,z$), sum to $360$ degrees. Furthermore the perimeter can be expressed as the following:
$$2(\sin(x/2)+\sin(y/2)+\sin(z/2))$$.
Since $\sin$ is concave on the interval $[0,90]$ and that $x,y,z \le 180$ we know by Jensen's inequality that the maximal value of this expression occurs when $x=y=z=120$, implying the maximum perimeter is $6\sin(60) = 3\sqrt{3}$.
An attempt to extend to the 3d case:
My hope was that I could find a way to extend this line of reasoning to three dimensions via the following steps:
Let $\theta_1,...,\theta_6$ be the angles formed at the center of the sphere which are in plane with a particular edge of the tetrahedron: to be clear, let $O$ be the center of the sphere, then $\theta_j$ is something like $\angle AOB$. The hope is that they add to an invariant, call it $I$.
It isn't hard to see that a tetrahedron that maximizes the sum of the side lengths must contain in its interior the center of the sphere. We get then that $0< \theta_j \le 180$.
The sum of the side lengths of the tetrahedron is $2(\sum_{j=1}^6\sin(\theta_j/2))$. So, again, we would apply Jensen's inequality to get $2\sum_{j=1}^6 \sin(\theta_j/2)\le 2 \cdot 6\sin(I/12)$.
Equality would only hold if $\theta_1 =...= \theta_6$ implying regularity. Furthermore, the hope is that $12\sin(I/12) =\frac{6\cdot2^{2/3}}{3^{1/2}}$.
Note from the above we can derive what $I$ should be:
$$\sin(I/12)=\frac{1}{2^{1/3}3^{1/2}}= (\frac{1}{108^{1/6}}) \Rightarrow I=12\sin^{-1}\big(\frac{1}{108^{1/6}}\big)$$.
The main issue is that I can't find a justification that the these angles should add to this invariant. Perhaps this approach was doomed from the start? I am uncertain if the supposed invariant is even invariant because I don't have a great way of constructing tetrahedra to test it. I was hoping someone would have insight on this.
My main takeaway question:
Given a tetrahedron with constituent points $A,B,C,D$ lying on a sphere with the center, $O$, which itself lies in the interior of the tetrahedron, is the sum $$\angle AOB + \angle AOC + \angle AOD + \angle BOC + \angle BOD + \angle COD$$ constant across all such tetrahedra, and how would we find it if so?
Another way to ask the same question after some research:
Given a tetrahedron, $A,B,C,D$ with circumcenter, $O$, lying in its interior, is the sum:
$$\angle AOB + \angle AOC + \angle AOD + \angle BOC + \angle BOD + \angle COD$$
constant across all such polyhedra?