Lei Lei wrote: Let point $A_1$ and $A_2$ be foot of perpendicular from $A_1$ onto $AB$ and $AC.$ This is clearly a typo, thus, for convenience I'll denote the orthogonal projections of $A'$ on $AC,AB$ as $Y,Z$ and $X$ is the orthogonal projection of $A'$ onto $BC.$ Circumcenter $O$ of $\triangle ABC$ is the intersection of the internal bisector $AA'$ of $\angle BA'C$ and the perpendicular bisector of $BC$ $\Longrightarrow$ $O$ lies on the circumcircle of $\triangle A'BC$ and it's the midpoint of its arc $BC.$ Hence, circumcircle $(O)$ of $\triangle ABC$ cuts $AA'$ at the incenter $K$ and A-excenter $A$ of $\triangle A'BC$ $\Longrightarrow$ $CA,BA$ are the external bisectors of $\angle BCA'$ and $\angle CBA'$ $\Longrightarrow$ orthogonal projections of $A'$ onto $CA,BA$ lie then on the A'-midline of $\triangle A'BC.$ Consequently, if ray $AH_A$ cuts $\odot(AYZ)$ at $D,$ then $D$ is the reflection of $H_A$ across $YZ,$ i.e. $H_A$ is the orthocenter of $\triangle AYZ \ (*).$ On the other hand, from cyclic quadrilaterals $YXCA'$ and $ZBXA',$ we have $\angle A'XY= \angle A'CY=\angle ACB \ , \ \angle A'XZ=\angle A'BZ=\angle ABC$ $\Longrightarrow \angle ZXY=\angle ACB+\angle ABC=\angle YA'Z$ Thus, $\odot(XYZ)$ is the reflection of $\odot(AYZ)$ across $YZ,$ hence $(*)$ implies that $H_A \in \odot(XYZ)$ and $AA'=2R_A.$ Let $AA',BB',CC'$ cut $BC,CA,AC$ at $P,Q,R.$ From $(A',P,K,A)=-1,$ we get $OA^2=R^2= OP \cdot OA'.$ $AA'=R+OA' \Longrightarrow 2R_A=R+OA'=R+\frac{R^2}{OP} \Longrightarrow 2R_A=R \cdot \frac{AP}{OP}$ By similar reasoning, we get $ \ : \ 2R_B=R \cdot \frac{BQ}{OQ} \ , \ 2R_C=R \cdot \frac{CR}{OR}$ $\Longrightarrow \frac{1}{R_A}+\frac{1}{R_B}+\frac{1}{R_C}=\frac{2}{R} \left ( \frac{OP}{AP}+\frac{OQ}{BQ}+\frac{OR}{CR} \right)$ $\frac{OP}{AP}+\frac{OQ}{BQ}+\frac{OR}{CR}=\frac{|\triangle BOC|}{|\triangle ABC|}+\frac{|\triangle COA|}{|\triangle ABC|}+\frac{|\triangle AOB|}{|\triangle ABC|}=\frac{|\triangle ABC|}{|\triangle ABC|}=1$ $\Longrightarrow \frac{1}{R_A}+\frac{1}{R_B}+\frac{1}{R_C}=\frac{2}{R}.$

Similar solution to above: Since $A'O$ the angle bisector of $\angle BA'C$ and $OB=OC$ it follows that $A' \in (BOC)$. Let $P$ be the second intersection of $AO$ and $(ABC)$ since $O$ is the midpoint of $arc$ $BOC$ and $OP=OB=OC=OA$ $\implies$ $P$ is the $incenter$ of $\triangle A'BC$ and $A$ is the $A-excenter$ of $\triangle A'BC$ so $PB \perp AA_1$ and $PC \perp AA_2$ so we have $\frac{AB}{AA_1}=\frac{AP}{AA'}=\frac{AC}{AA_2} \implies BC \parallel A_1A_2$ call this fact $(1)$. Let $H$ be the orthocenter of $\triangle ABC$ we shall prove that $\triangle BHC$ is homothetic to $\triangle A_1H_AA_2$ to prove that it is enought to prove $HC \parallel H_AA_2$ and then $BH \parallel H_AA_1$ is proven analogously which than implies $\triangle BHC$ is homothetic to $\triangle A_1H_AA_2$. To prove this we first do an angle chase (which i will skip) to get $\triangle AH_BH_A \sim \triangle ABA'$ and $\triangle AH_BH \sim \triangle ABP$ so we obtain $\frac{AB}{AA'}=\frac{AH_B}{AH_A}$ and $\frac{AP}{AB}=\frac{AH}{AH_B}$ multyplying and using $(1)$ we get $\frac{AC}{AA_2}=\frac{AH}{AH_A} $which proves our fact. Let $R(XYZ)$ be the radius of $(XYZ)$ We have $\frac{R(BCH)}{R(A_1A_2H_A)}=\frac{HC}{H_AA_2}$ $\implies$ $\frac{1}{R_A}=\frac{1}{R} \cdot \frac{AH}{AH_A}$ because $R(BCH)=R(ABC)$ Now define $x$, $y$, $z$ to be the areas of $\triangle HBC$ , $\triangle HCA$ and $\triangle HAB$, respectively. We easily get $\frac{AH}{AH_A}=\frac{y+z}{x+y+z}$ and cyclically. Finaly we get $\sum\limits_{cyc}{ \frac{1}{R_A}}= \sum\limits_{cyc}{ \frac{1}{R} \cdot \frac{AH}{AH_A}}=\frac{1}{R} \cdot \sum\limits_{cyc}{ \frac{AH}{AH_A}}=\frac{1}{R} \cdot \sum\limits_{cyc}{ \frac{y+z}{x+y+z}}=\frac{2}{R}$ $$Q.E.D$$