86. As shown in the figure, let the circumradius of $\triangle ABC$ be $R$, and the circumradius of $\triangle BOC$ be $R'$. It is easy to see that $\angle BAO = \angle ABO = 90^\circ - C$, $\angle CAO = \angle ACO = 90^\circ - B$, $\angle OBC = \angle OCB = \angle CA_1O = 90^\circ - A$, $\angle OA_2C = \angle ABC + \angle BAO = 90^\circ + B - C$, $\angle COA_1 = \angle CAO + \angle ACO = 180^\circ - 2B$, $\angle OCA_1 = 180^\circ - \angle COA_1 - \angle OA_1C = 180^\circ - (180^\circ - 2B) - (90^\circ - A) = 2B + A - 90^\circ = 90^\circ - C + B$. Applying the Law of Sines in $\triangle ABC$ and $\triangle BOC$, we get $BC = a = 2R \sin A = 2R \cdot \sin \angle BOC = 2R' \sin 2A$, so $2R' = \frac{R}{\cos A}$. Using the Law of Sines again, we have
$$\begin{array}{l} OA_1 = 2R' \sin \angle OCA_1 = \frac{R \sin (90^\circ - C + B)}{\cos A} = \frac{R \cos (B - C)}{\cos A} \\ AA_1 = AO + OA_1 = R + \frac{R \cos (B - C)}{\cos A} = R \left(1 + \frac{\cos (B - C)}{\cos A}\right) \\ \frac{AA_1}{OA_1} = \frac{R \left(1 + \frac{\cos (B - C)}{\cos A}\right)}{\frac{R \cos (B - C)}{\cos A}} = \frac{\cos A}{\cos (B - C)} + 1 \end{array}$$

Similarly,
$$\begin{array}{c} \frac{BB_1}{OB_1} = \frac{\cos B}{\cos (C - A)} + 1 \\ \frac{CC_1}{OC_1} = \frac{\cos C}{\cos (A - B)} + 1 \\ \frac{AA_1}{OA_1} + \frac{BB_1}{OB_1} + \frac{CC_1}{OC_1} \geq \frac{9}{2} \Leftrightarrow \frac{\cos A}{\cos (B - C)} + \frac{\cos B}{\cos (C - A)} + \frac{\cos C}{\cos (A - B)} \geq \frac{3}{2} \end{array}$$

And
$$\frac{\cos A}{\cos (B - C)} = \frac{2 \sin A \cos A}{2 \sin (B + C) \cos (B - C)} = \frac{\sin 2A}{\sin 2B + \sin 2C}$$

Similarly,
$$\begin{array}{l} \frac{\cos B}{\cos (C - A)} = \frac{\sin 2B}{\sin 2C + \sin 2A} \\ \frac{\cos C}{\cos (A - B)} = \frac{\sin 2C}{\sin 2A + \sin 2B} \end{array}$$

By Nesbitt's inequality $\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} \geq \frac{3}{2}$, we get
$$\frac{\sin 2A}{\sin 2B + \sin 2C} + \frac{\sin 2B}{\sin 2C + \sin 2A} + \frac{\sin 2C}{\sin 2A + \sin 2B} \geq \frac{3}{2}$$