Let $A,B,C$ be the angles of a triangle whose vertices are uniformly random on a circle. My experimental data suggests that
$$ P\left(\frac{1}{A} + \frac{1}{B} > \frac{1}{C}\right) = \frac{2}{5} + \frac{8}{5\sqrt{5}}\sinh^{-1}\left({\frac{1}{2}}\right) \approx 0.74432 $$
Can this be proved or disproved?
Related problem: Probability that the reciprocal triangle inequality $\frac{1}{a} + \frac{1}{b} \ge \frac{1}{c}$ holds.
We have $$A+B+C=\pi$$ So the probability in question is equivalent to $$P=\frac{\int_{0}^{\pi}\int_{0}^{\pi-A}\mathbf{1}_{\frac{1}{A}+\frac{1}{B}>\frac{1}{\pi-A-B}}\,\mathrm dB\,\mathrm dA}{\int_{0}^{\pi}\int_{0}^{\pi-A}1\,\mathrm dB\,\mathrm dA}=\frac{2}{\pi^2}\int_{0}^{\pi}\int_{0}^{\pi-A}\mathbf{1}_{\frac{1}{A}+\frac{1}{B}>\frac{1}{\pi-A-B}}\,\mathrm dB\,\mathrm dA$$ Now we analyze the indicator function: $$\frac{1}{A}+\frac{1}{B}>\frac{1}{\pi-A-B}$$ $$\pi A+\pi B-A^2-2AB-B^2>AB$$ $$A^2+3AB+B^2-\pi A-\pi B<0$$ $$B^2+B(3A-\pi)+A^2-\pi A<0$$ $$\frac{\pi-3A-\sqrt{(\pi-3A)^2-4\left(A^2-\pi A\right)}}{2}<B<\frac{\pi-3A+\sqrt{(\pi-3A)^2-4\left(A^2-\pi A\right)}}{2}$$ $$\frac{\pi-3A-\sqrt{5A^2-2\pi A+\pi^2}}{2}<B<\frac{\pi-3A+\sqrt{5A^2-2\pi A+\pi^2}}{2}$$ We can easily show that $\frac{\pi-3A-\sqrt{5A^2-2\pi A+\pi^2}}{2}\le 0$ and $\frac{\pi-3A+\sqrt{5A^2-2\pi A+\pi^2}}{2}\le\pi-A$. So the lower bound must be improved to $0$, while the upper bound stays the same.
Therefore $$P=\frac{2}{\pi^2}\int_{0}^{\pi}\frac{\pi-3A+\sqrt{5A^2-2\pi A+\pi^2}}{2}\,\mathrm dA$$ The remaining integral is particularly easy to deal with.
