China Mathematical Olympiad

Firstly we claim that $A', B, O, C$ are concyclic points. Otherwise, extend $AO$ to intersect the circumcircle of $\triangle BOC$ at point $P$ which is different from $A'$. We get
$$\angle BPA = \angle BCO = \angle CBO = \angle CPA.$$
Then $\triangle PA'C \cong \triangle PA'B$, and $A'B = A'C$. So $AB = AC$ and that is a contradiction since $\triangle ABC$ is not isosceles. So
$$\angle BCA' = \angle BOA' = 180^\circ - 2\angle C,$$
and $\angle A'CA_1 = \angle C$.
Further, we have
$$\frac{H_AA_1}{AC} = \sin \angle C = \cos \angle A_2AA' = \frac{AA_2}{AA'}$$
and
$$\angle A'AC = \angle A_2AH_A = \frac{\pi}{2} - \angle B.$$
So $\triangle A_2AH_A \sim \triangle A'AC$. In the same way, $\triangle A_1H_A A \sim \triangle A'BA$. Then $\angle A_2H_A A = \angle ACA'$ and $\angle A_1H_A A = \angle ABA'$. Consequently,
$$\begin{align*} \angle A_1 H_A A_2 &= 2\pi - \angle A_2 H_A A - \angle A_1 H_A A \\ &= 2\pi - \angle ACA' - \angle ABA' \\ &= \angle A + 2\left(\frac{\pi}{2} - \angle A\right) \\ &= \pi - \angle A. \end{align*}$$
We get
$$\frac{R}{R_A} = \frac{\frac{R}{A_1 A_2}}{\frac{2 \sin \angle A_1 H_A A_2}{\sin \angle A}} = \frac{2R}{A_1 A_2} = \frac{2R}{AA'}; \quad \textcircled{1}$$
The last equality holds since $A, A_2, A', A_1$ lie on the same circle with $AA'$ as the diameter.
Now, draw $AA'' \perp A'C$ with point $A''$ on line $A'C$. Since $\angle ACA'' = \angle A'CA_1 = \angle C$, we have $AA'' = AH_A$. Then
$$\begin{align*} AA' &= \frac{AA''}{\sin \angle AA'C} = \frac{AH_A}{\sin(90^\circ - \angle A)} \\ &= \frac{AH_A}{\cos \angle A} = \frac{2S_{\triangle ABC}}{BC \cos \angle A}. \quad \textcircled{2} \end{align*}$$
From ①, ② we get
$$\begin{align*} \frac{1}{R_A} &= \frac{BC \cos \angle A}{S_{\triangle ABC}} = \frac{\cos \angle A}{R \sin \angle B \sin \angle C} \\ &= \frac{1}{R} (1 - \cot \angle B \cot \angle C). \end{align*}$$
In the same way
$$\frac{1}{R_B} = \frac{1}{R}(1 - \cot \angle C \cot \angle A)$$
and
$$\frac{1}{R_C} = \frac{1}{R}(1 - \cot \angle A \cot \angle B).$$
Notice that
$$\cot \angle A \cot \angle B + \cot \angle B \cot \angle C + \cot \angle C \cot \angle A = 1.$$
We then have
$$\frac{1}{R_A} + \frac{1}{R_B} + \frac{1}{R_C} = \frac{2}{R}.$$