inscribed vs. circumscribed randomly generated similar triangles: which one has more ‘excess’ area on average? by how much Obtain two similar triangles with uniformly-randomly chosen angle measures, assigning the larger one an area of 1 and the smaller one an area such that its circumcircle has an area of 1. Which area on average would you expect to be greater: the area inside the larger triangle but outside its incircle or the area inside the larger circle but outside its inscribed triangle? (And what are these average areas? Minimum and maximum?)
 A: Not an answer, just a walk-through of the formula reduction from a comment ...

OP is comparing the ratio of excess triangle area ($\Delta$) to incircle area ($I$) to the ratio of excess circumcircle area ($K$) to triangle area, so we can write
$$\begin{align}
\require{cancel}
\frac{\Delta-I}{\Delta}&\;\lesseqgtr\;\frac{K-\Delta}{K} \tag1\\[4pt]
\text{K}\cdot(\Delta-I) &\;\lesseqgtr\; \Delta\cdot(K-\Delta) \tag2 \\[4pt]
\cancel{K\Delta}-KI &\;\lesseqgtr\; \cancel{\Delta K}-\Delta^2 \tag3\\[4pt]
\Delta^2 &\;\lesseqgtr\; KI =  \pi R^2\cdot\pi r^2=\left(\pi rR\right)^2 \tag4\\[4pt]
\Delta &\;\lesseqgtr\; \pi r R \tag5
\end{align}$$
where $r$ and $R$ are inradius and circumradius. Note that the order of the inequalities has been (and will be) preserved; the left-hand side is smaller than (or equal to, or larger than) the right-hand side in $(1)$ whenever this is true in $(2)$, $(3)$, $(4)$, etc.
From here, we can invoke some relations between the quantities, and a little trig. (See, for instance, MathWorld's "Inradius", "Law of Sines", and "Double-Angle Formulas" entries.)
$$\begin{align}
r &= 4R\sin\tfrac12A\sin\tfrac12B\sin\tfrac12C \tag6 \\[4pt]
\Delta &= \tfrac12ab\sin C =\tfrac12\cdot 2R\sin A\cdot 2R\sin B\cdot \sin C  \tag7 \\[4pt] 
&= 2 R^2 \cdot 2\sin\tfrac12A\cos\tfrac12A\cdot 2\sin\tfrac12B\cos\tfrac12B\cdot 2\sin\tfrac12 C\cos\tfrac12C \tag8 \\[4pt]
&= 4 R \sin\tfrac12A\sin\tfrac12B\sin\tfrac12C \cdot 4R\cos\tfrac12A\cos\tfrac12B\cos\tfrac12C \tag9 \\[4pt]
&= 4 r R \cos\tfrac12A\cos\tfrac12B\cos\tfrac12C \tag{10}
\end{align}$$
Substituting $(10)$ into $(5)$, and dividing-through by $4rR$, yields

$$\cos\tfrac12A\cos\tfrac12B\cos\tfrac12C \;\lesseqgtr\; \frac\pi4\tag{$\star$}$$

as claimed. $\square$
