= Solution
Assume distinct positive separations, and first label them so $A<B$. Let $x=R_0^6$, $a=A^6$ and $b=B^6$. The positive contrast is
$$
\Delta E=E(A)-E(B)=\frac{x}{x+a}-\frac{x}{x+b}
=\frac{x(b-a)}{(x+a)(x+b)}.
$$
Its derivative has a particularly simple sign:
$$
\frac{d\Delta E}{dx}=\frac{(b-a)(ab-x^2)}{(x+a)^2(x+b)^2}.
$$
It is positive below $x=\sqrt{ab}$ and negative above, so this is the global maximum. Both end limits $x\to0$ and $x\to\infty$ give zero contrast. Thus
$$
\boxed{R_{0,\rm opt}=\sqrt{AB},\qquad
\Delta E_{\max}=\frac{B^3-A^3}{B^3+A^3}.}
$$
To recover the printed expression, factor its sixth-power radicand:
$$
\frac{A^6-(AB)^3}{(A/B)^3-1}
=\frac{A^3(A^3-B^3)}{(A^3-B^3)/B^3}=A^3B^3.
$$
Its sixth root is indeed $\sqrt{AB}$. The cancellation requires $A\ne B$. If $A=B$, every choice gives zero contrast and there is no unique optimum; the printed expression then has a removable algebraic $0/0$ without a unique maximization conclusion. For unordered $A,B$, maximize \b[absolute] contrast and replace the numerator of the maximum by $|B^3-A^3|$. A negative signed difference for $A>B$ is minimized, not maximized, by the same interior point. These qualifications are needed to make the statement precise.
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