= Solution
To find the <optimal Förster distance for two-state contrast>, take positive distinct distances $A,B$ and put $x=R_0^6>0$. The magnitude of the efficiency change is
$$
\Delta E=\left|\frac{x}{x+A^6}-\frac{x}{x+B^6}\right|
=\frac{x|B^6-A^6|}{(x+A^6)(x+B^6)}.
$$
Its <derivative> has the sign of $A^6B^6-x^2$. It rises from zero, has a single maximum at $x=(AB)^3$, then falls back to zero. Hence
$$
\boxed{R_{0,\rm opt}=\sqrt{AB}.}
$$
The printed form is equivalent, since
$$
\frac{A^6-(AB)^3}{(A/B)^3-1}
=\frac{A^3(A^3-B^3)}{(A^3-B^3)/B^3}=A^3B^3.
$$
At this optimum the two efficiencies sum to one and the maximal contrast is $|B^3-A^3|/(B^3+A^3)$. If $A=B$, there is no distance change and the contrast is zero for every $R_0$; the source's unsimplified expression is then $0/0$ and does not specify a unique optimum. For example $A=B=5\,\mathrm{nm}$ yields zero contrast regardless of the fluorophore pair. The geometrical-mean result is a design criterion for distinct distances within the assumed fixed-orientation/photophysics <FRET> law, not a correction for dye-orientation changes.
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