I found this curious geometric probability expeerimentally. Can this be proved or disproved?
Let $a$ be the smallest side of a triangle whose inradius is $r$ and vertices are uniformly random on circle of radius $R$. Experimental data for $10^9$ simulations show that the probability
$$ P\left(Rr > \frac{a^2}{2}\right) = \frac{1}{2}. $$
Consider the opposite probability. $$Rr\le \frac{a^2}{2}$$ $$\implies\frac{abc}{4\Delta}\frac{2\Delta}{a+b+c}\le \frac{a^2}{2}$$ $$\implies bc\le a(a+b+c)$$ $$\implies \frac{a}{b}\cdot\frac{a}{c}+\frac{a}{c}+\frac{a}{b}\ge 1$$ Using argument similar to this answer, let $$a=2\sin\alpha,b=2\sin(\alpha+\beta),c=2\sin\beta$$ With $$\alpha,\beta>0;\alpha+\beta<\pi$$ And consider $$s=\frac{c}{a}=\frac{\sin\beta}{\sin\alpha},t=\frac{b}{a}=\frac{\sin(\alpha+\beta)}{\sin\alpha}$$ The mapping $(\alpha,\beta)\mapsto(s,t)$ is bijective onto the set of pairs $(s,t)\in\mathbb{R}^2$ satisfying $$s,t>0;s+t>1>|s-t|$$ And the Jacobian determinant of the mapping is $st$, the condition $\min(a,b,c)=a$ has Lebesgue measure $\pi^2/6$
Using the change of variable $(\alpha,\beta)\mapsto(s,t)$, we get $$P\left(Rr\le \frac{a^2}{2}\mid \min(a,b,c)=a\right)=\frac{6}{\pi^2}\int_{\Omega}\frac{\mathrm ds\,\mathrm dt}{st}$$ With $$\Omega=\left\{(s,t)\in\mathbb{R}^2:s,t\ge 1;\frac{1}{st}+\frac{1}{s}+\frac{1}{t}\ge 1\right\}$$ Consider the condition $$\frac{1}{st}+\frac{1}{s}+\frac{1}{t}\ge 1$$ $$\implies \frac{1}{t}\left(\frac{1}{s}+1\right)\ge 1-\frac{1}{s}$$ $$\implies t\le \frac{1+1/s}{1-1/s}=\frac{s+1}{s-1}$$ However since we have $1>|s-t|$ then $s+1>t>s-1$, and: $$\Omega=\left\{(s,t)\in\mathbb{R}^2:1\le s<2;1\le t<s+1\right\}\\\cup\left\{(s,t)\in\mathbb{R}^2:2\le s<3;s-1<t\le \frac{s+1}{s-1}\right\}$$ Apply Fubini's theorem: $$\frac{6}{\pi^2}\int_{\Omega}\frac{\mathrm ds\,\mathrm dt}{st}=\frac{6}{\pi^2}\left(\int_{1}^{2}\int_{1}^{s+1}\frac{\mathrm dt}{t}\,\frac{\mathrm ds}{s}+\int_{2}^{3}\int_{s-1}^{(s+1)/(s-1)}\frac{\mathrm dt}{t}\,\frac{\mathrm ds}{s}\right)$$ $$=\frac{6}{\pi^2}\left(\int_{1}^{2}\frac{\ln(s+1)}{s}\,\mathrm ds+\int_{2}^{3}\frac{\ln(s+1)-2\ln(s-1)}{s}\,\mathrm ds\right)$$ $$=\frac{6}{\pi^2}\left(\int_{1}^{3}\frac{\ln(s+1)}{s}\,\mathrm ds-2\int_{2}^{3}\frac{\ln(s-1)}{s}\,\mathrm ds\right)$$ $$=\frac{6}{\pi^2}\left(\int_{2}^{4}\frac{\ln s}{s-1}\,\mathrm ds-2\int_{1}^{2}\frac{\ln s}{s+1}\,\mathrm ds\right)$$ We have: $$\int\frac{\ln x}{x-1}\,\mathrm dx=-\operatorname{Li}_2(1-x)+C$$ $$\int\frac{\ln x}{x+1}\,\mathrm dx=\operatorname{Li}_2(-x)+\ln x\ln(x+1)+C$$ $$\implies P\left(Rr\le \frac{a^2}{2}\mid \min(a,b,c)=a\right)$$ $$=\frac{6}{\pi^2}\left(-\operatorname{Li}_2(-3)+\operatorname{Li}_2(-1)-2\operatorname{Li}_2(-2)+2\operatorname{Li}_2(-1)-2\ln 2\ln 3\right)$$ $$=\frac{6}{\pi^2}\left(-\operatorname{Li}_2(-3)-2\operatorname{Li}_2(-2)+3\operatorname{Li}_2(-1)-2\ln 2\ln 3\right)$$ To prove this evaluates to $1/2$, first we apply the identity: $$\operatorname{Li}_2(z)+\operatorname{Li}_2\left(\frac{1}{z}\right)=-\frac{\pi^2}{6}-\frac{\ln^2(-z)}{2}$$ $$\implies \operatorname{Li}_2(-3)+\operatorname{Li}_2\left(-\frac{1}{3}\right)=-\frac{\pi^2}{6}-\frac{\ln^2 3}{2},$$ $$\operatorname{Li}_2(-2)+\operatorname{Li}_2\left(-\frac{1}{2}\right)=-\frac{\pi^2}{6}-\frac{\ln^2 2}{2}$$ And the following identities by Ramanujan: $$\operatorname{Li}_2\left(-\frac{1}{3}\right)-\frac{1}{3}\operatorname{Li}_2\left(\frac{1}{9}\right)=-\frac{\pi^2}{18}+\frac{\ln^2 3}{6},$$ $$\operatorname{Li}_2\left(-\frac{1}{2}\right)+\frac{1}{6}\operatorname{Li}_2\left(\frac{1}{9}\right)=-\frac{\pi^2}{18}+\ln 2\ln 3-\frac{\ln^2 2}{2}-\frac{\ln^2 3}{3}$$ $$\implies \operatorname{Li}_2\left(-\frac{1}{3}\right)+2\operatorname{Li}_2\left(-\frac{1}{2}\right)=-\frac{\pi^2}{6}+2\ln 2\ln 3-\ln^2 2-\frac{\ln^2 3}{2}$$ $$\implies \operatorname{Li}_2(-3)+2\operatorname{Li}_2(-2)=-\frac{\pi^2}{2}-\ln^2 2-\frac{\ln^2 3}{2}+\frac{\pi^2}{6}-2\ln 2\ln 3+\ln^2 2+\frac{\ln^2 3}{2}$$ $$=-\frac{\pi^2}{3}-2\ln 2\ln 3$$ $$\implies -\operatorname{Li}_2(-3)-2\operatorname{Li}_2(-2)+3\operatorname{Li}_2(-1)-2\ln 2\ln 3=\frac{\pi^2}{3}-\frac{\pi^2}{4}=\frac{\pi^2}{12}$$ $$\implies P\left(Rr\le \frac{a^2}{2}\mid \min(a,b,c)=a\right)=\frac{6}{\pi^2}\cdot\frac{\pi^2}{12}=\frac{1}{2}$$ Thus $$\boxed{P\left(Rr>\frac{a^2}{2}\mid \min(a,b,c)=a\right)=\frac{1}{2}}$$
Long comment.
Let:
It can be shown that $P_1=\frac14$. (In the link, it is assumed that $\frac{a}{b}>1$ and $P\left(\frac{ac}{b}>a+b\right)$ is shown to be $\frac12$. Now if $\frac{a}{b}<1$ then $P\left(\frac{ac}{b}>a+b\right)=0$ by the triangle inequality. So $P_1=\frac14$.)
A simulation suggests that $P_2=\frac16$. I suspect that the method in the above link can be used to prove this, because $P_1$ and $P_2$ are similar in form.
It can be shown that $P_3=3P_2$, by the following argument. If $b$ is not the triangle's smallest side length, then assuming (without loss of generality) that $a$ is the triangle's smallest side length, we must have $\frac{ac}{b}<c<a+b+c$, that is, $P\left(\frac{ac}{b}>a+b+c\right)=0$. So $P_3=3P_2$.
Euclid's comment to the OP shows that $P_3$ is equivalent to the probability in the OP (note that I'm calling the smallest side length $b$ instead of $a$).
This would show that the probability in the OP is indeed $\frac12$.
