Could somebody compute the following two sums as a function of known things?(If yes the problem is over ) So please compute $\angle{AP_{A}I}+\angle{AQ_{A}I}$ and $\angle{AP_{A}I}-\angle{AQ_{A}I}$

Hey nice problem We can show $P_{A}Q_{A}+P_{B}Q_{B}+P_{C}Q_{C}\le\frac{3}{2}R$ Indeed we can show $P_{A}Q_{A}=\frac{a^{2}(b+c-a)}{2S}$ thus $\sum P_{A}Q_{A}\le\frac{3}{2}R\Leftrightarrow \sum\frac{a^{2}(b+c-a)}{2S}\le \frac{3}{2}R\Leftrightarrow a^{3}+b^{3}+c^{3}+3abc\ge \sum (b+c)a$ Trivial by Schur! But I think the hard problem here is show $P_{A}Q_{A}=\frac{a^{2}(b+c-a)}{2S}$ I have a solution by calculation can anybody show a nice proof?

gemath wrote: But I think the hard problem here is show $P_{A}Q_{A}=\frac{a^{2}(b+c-a)}{2S}$ I have a solution by calculation can anybody show a nice proof? Could you post your way ?Thank you

Ok here is my solution for that indentity for simple I change notation $P_{A},Q_{A}$ into $P,Q$ We easily show $IP+OP=R+r,IQ+OQ=R-r,PQ=OP-OQ$ here we only use triangle $AIO$ noate that $P\in AO,OA=R$ and $IP+OP=R+r$ that's all for we calculate now let $\frac{OP}{OA}=x=\frac{OP}{R}\Rightarrow \frac{PA}{R}=1-x$ now by Stewart's theorem in triangle $IOA$ we have $\frac{1}{x}IO^{2}+\frac{1}{1-x}IA^{2}=(\frac{1}{x}+\frac{1}{1-x})IP^{2}+R^{2}$ and using $IP=R+r-OP=(1-x)R+r,OI^{2}=R^{2}-2Rr$ we easily show $OP=Rx=R\frac{4Rr+r^{2}}{IA^{2}+4Rr}$ similarly for $Q\in OA$ let $\frac{OQ}{OA}=y$ but note that $IQ=(1-y)R-r$ we have $OQ=Ry=R\frac{r^{2}}{IA^{2}}$ now using $PQ=OP-OQ=R\frac{IA^{2}-r^{2}}{IA^{2}+4Rr}$ now note that $IA^{2}=\frac{(p-a)^{2}}{\cos^{2}\frac{A}{2}}=\frac{sbc}{s-a}$ with $s$ is semiperimeter and $4Rr=\frac{abc}{p}$ we easily seen my indentity.

Many thanks .Your way is straightforward . Now could you find something for the angles I wrote above ?

My proof is basically the same of yours, gemath! Very Nice problem. Now, i'm trying to find one solution using inversion!

:spider: Sorry Gemath, you commit some small mistakes ! Lemma (Gemath). Let $ABC$ be an acute triangle with the incircle $\omega =C(I,r)$ and the circumcircle $\rho =C(O,R)$. The circles $C_{1}=C(P,r_{1})$ and $C_{2}=(Q,r_{2})$ are tangent internally to the circle $\rho$ in the same point $A$ The circle $w$ is tangent externally to the circle $C_{1}$ and is tangent internally to the circle $C_{2}$. Prove that $\boxed{PQ=\frac{a^{2}(p-a)}{4S}}$, where $2p=a+b+c$ and $S\equiv [ABC]$- the area of $\triangle ABC$. Proof. Prove easily that $\{\begin{array}{ccccc}IP=r+r_{1}& , & IQ=r_{2}-r\\\\ PO=R-r_{1}& , & PA=r_{1}\\\\ QO=R-r_{2}& , & QA=r_{2}\\\\ OA=R & , & \boxed{PQ=r_{2}-r_{1}}\end{array}$. The relations $\{\begin{array}{ccc}IO^{2}=R(R-2r) & , & IA^{2}=\frac{bc(p-a)}{p}\\\\ IA^{2}-r^{2}=(p-a)^{2}& , & IA^{2}+4Rr=bc\\\\ p(p-a)+(p-b)(p-c)=bc\end{array}$ are well-known. Denote $IO=m$, $IA=n$ and apply the Stewart's theorem in $\triangle OIA$ for the cevian-rays $[IP$ and $[IQ$ : $\{\begin{array}{c}m^{2}r_{1}^{2}+n^{2}(R-r_{1})=R(r+r_{1})^{2}+Rr_{1}(R-r_{1})\\\\ m^{2}r_{2}+n^{2}(R-r_{2})=R(r_{2}-r)^{2}+Rr_{2}(R-r_{2})\end{array}\|$ $\implies$ $r_{1}(R^{2}+2Rr+n^{2}-m^{2})=R(n^{2}-r^{2})=r_{2}(R^{2}-2Rr+n^{2}-m^{2})$ $\implies$ $r_{1}(R^{2}+2Rr+bc-4Rr-R^{2}+2Rr)=$ $R(p-a)^{2}=$ $r_{2}(R^{2}-2Rr+bc-4Rr-R^{2}+2Rr)$ $\implies$ $r_{1}bc=R(p-a)^{2}=r_{2}(bc-4Rr)$ $\implies$ $PQ=r_{2}-r_{1}=$ $\frac{4R^{2}r(p-a)^{2}}{bc(bc-4Rr)}=$ $\frac{4R^{2}pr(p-a)^{2}}{b^{2}c^{2}(p-a)}=$ $\frac{4R^{2}pra^{2}(p-a)^{2}}{16R^{2}p^{2}r^{2}(p-a)}$ $\implies$ $\boxed{\ PQ=\frac{a^{2}(p-a)}{4S}\ }$. Remark. $8\cdot \prod PQ=\frac{8(abc)^{2}(p-a)(p-b)(p-c)}{(4S)^{3}}=\frac{8(4RS)^{2}Sr}{(4S)^{3}}=$ $R^{2}\cdot 2r$ $\implies$ $\{\begin{array}{c}\boxed{\ r\le \sqrt [3]{\prod PQ}\le \frac{R}{2}\ }\\\\ \boxed{\ r\le\frac{1}{3}\cdot \sum PQ\le \frac{R^{2}}{4r}\ }\end{array}$

I think my result is true $P_{A}Q_{A}=\frac{a^{2}(b+c-a)}{2S}=\frac{a^{2}(s-a)}{4S}=\frac{a^{2}}{4r_{a}}$ but maybe I made some mistake when I wrote this problem here, thanks for your remark Mr Virgil. An I think the inequality $\sum P_{A}Q_{A}\le \frac{3}{2}R$ is true, it is Schur? and note that we can get $\sum \frac{P_{A}Q_{A}}{r_{a}}\ge \frac{9}{16}$ ($r_{a},r_{b},r_{c}$ are exradius of $ABC$) from Iran inequality.

gemath wrote: I think my result is true $\boxed{P_{A}Q_{A}=\frac{a^{2}(b+c-a)}{2S}=\frac{a^{2}(s-a)}{4S}}=\frac{a^{2}}{4r_{a}}$ but maybe I made some mistake when I wrote this problem here, thanks for your remark Mr Virgil. An I think the inequality $\sum P_{A}Q_{A}\le \frac{3}{2}R$ is true, it is Schur? and note that we can get $\sum \frac{P_{A}Q_{A}}{r_{a}}\ge \frac{9}{16}$ ($r_{a},r_{b},r_{c}$ are exradius of $ABC$) from Iran inequality. Gemath, if $2s=a+b+c$, then the relation $\frac{a^{2}(b+c-a)}{2S}=\frac{a^{2}(s-a)}{4S}$ is falsely. Correctly, $P_{A}Q_{A}=\frac{a^{2}(s-a)}{4S}$. I used no Schur in the my proof.

Ok that's right I made a small mistake , but I think solution of stronger by Schur is true? And moreover here we don't need acute triangle my solution is true for all triangle ?

This problem was proposed by Sung-Yoon Kim (MIT).

Let $ ABC$ be an acute triangle with the incircle $ \omega = C(I,r)$ and the circumcircle $ \rho = C(O,R)$. The circles $ C_{1}= C(P,r_{1})$ and $ C_{2}= (Q,r_{2})$ are tangent internally to the circle $ \rho$ in the same point $ A$ The circle $ w$ is tangent externally to the circle $ C_{1}$ and is tangent internally to the circle $ C_{2}$. Prove that $ \boxed{PQ =\frac{a^{2}(p-a)}{4S}}$, where $ 2p = a+b+c$ and $ S\equiv [ABC]$- the area of $ \triangle ABC$. Prove easily that $ \{\begin{array}{ccccc}IP = r+r_{1}& , & IQ = r_{2}-r\\ \\ PO = R-r_{1}& , & PA = r_{1}\\ \\ QO = R-r_{2}& , & QA = r_{2}\\ \\ OA = R & , &\boxed{PQ = r_{2}-r_{1}}\end{array}$. The relations $ \{\begin{array}{ccc}IO^{2}= R(R-2r) & , & IA^{2}=\frac{bc(p-a)}{p}=\frac{4Rr(p-a)}{a}\\ \\ IA^{2}-r^{2}= (p-a)^{2}& , & IA^{2}+4Rr = bc\\ \\ p(p-a)+(p-b)(p-c) = bc.\end{array}$ are well-known. Proof I. Denote $ IO = m$, $ IA = n$. Apply the Stewart's theorem in $ \triangle OIA$ for the cevian-rays $ [IP$ and $ [IQ$ : $ \{\begin{array}{c}m^{2}r_{1}^{2}+n^{2}(R-r_{1}) = R(r+r_{1})^{2}+Rr_{1}(R-r_{1})\\ \\ m^{2}r_{2}+n^{2}(R-r_{2}) = R(r_{2}-r)^{2}+Rr_{2}(R-r_{2})\end{array}\|$ $ \implies$ $ r_{1}(R^{2}+2Rr+n^{2}-m^{2}) = R(n^{2}-r^{2}) = r_{2}(R^{2}-2Rr+n^{2}-m^{2})$ $ \implies$ $ r_{1}(R^{2}+2Rr+bc-4Rr-R^{2}+2Rr) =$ $ R(p-a)^{2}=$ $ r_{2}(R^{2}-2Rr+bc-4Rr-R^{2}+2Rr)$ $ \implies$ $ r_{1}bc = R(p-a)^{2}= r_{2}(bc-4Rr)$ $ \implies$ $ PQ = r_{2}-r_{1}=$ $ \frac{4R^{2}r(p-a)^{2}}{bc(bc-4Rr)}=$ $ \frac{4R^{2}pr(p-a)^{2}}{b^{2}c^{2}(p-a)}=$ $ \frac{4R^{2}pra^{2}(p-a)^{2}}{16R^{2}p^{2}r^{2}(p-a)}$ $ \implies$ $ \boxed{\ PQ =\frac{a^{2}(p-a)}{4S}\ }$. Proof II. Apply the Pythagoras' theorem in the triangles : $ \triangle\ IOP\blacktriangleright$ $ (r+r_{1})^{2}= (R-r_{1})^{2}+(R^{2}-2Rr)-2(R-r_{1})\cdot$ $ OI\cdot\frac{(R^{2}-2Rr)+R^{2}-\frac{4Rr(p-a)}{a}}{2R\cdot IO}\implies$ $ \boxed{r_{1}=\frac{R(p-a)^{2}}{bc}}\ .$ $ \triangle\ IOQ\blacktriangleright$ $ (r_{2}-r)^{2}= (R-r_{2})^{2}+(R^{2}-2Rr)-2(R-r_{2})\cdot$ $ IO\cdot\frac{(R^{2}-2Rr)+R^{2}-\frac{4Rr(p-a)}{a}}{2R\cdot IO}\implies$ $ \boxed{r_{2}=\frac{Rp(p-a)}{bc}}\ .$ Therefore, $ \frac{r_{1}}{p-a}=\frac{r_{2}}{p}=\frac{R(p-a)}{bc}=\frac{PQ}{a}$, i.e. $ \boxed{PQ =\frac{aR(p-a)}{bc}}=\frac{a^{2}(p-a)}{4S}$. Observe that $ \frac{r_{1}}{r}=\frac{r_{2}}{r_{a}}$, where $ r_{a}$ is the $ A$- exinradius of $ \triangle ABC$. Remark. Prove easily that two more interesting relations : $ \boxed{\sum PQ = R+r\ }$ and $ r_{1}= R-\frac{a(4R+r)}{4p}$ , $ \boxed{r_{2}=\frac{ r_{b}+r_{c}}{4}}$.

Another solution: if $ \angle BAC = 2A, \angle ABC = 2B \angle ACB = 2C$, then it's clear that $ \angle IAO = B-C$. It's also clear, that if $ O$ is the circumcentre of $ ABC$, that $ A, P_{A}, Q_{A}, O$ are collinear, and $ IP_{A} = AP_{A}+r$. But using the Cosine Rule on $ IAP_{A}$, it's clear that $ (IP_{A})^2 = IA^2 + AP_{A}^2 - 2(IA)(AP_{A}) cos (B-C) = AP_{A}^2 + r^2 + 2(AP_{A})r$; now using the fact that $ IA^2 - r^2 = (\frac {b+c-a}{2})^2$, we have $ AP_{A} = \frac {(b+c-a)^2} {8(IA cos(B-C)+r)}$. Similarly, $ AQ_{A} = \frac {(b+c-a)^2} {8(IA cos(B-C) - r)}$ Using the fact that $ IA = \frac {r}{sin a}$, gives us that $ P_{A}Q_{A}$ = $ AP_{A} - AQ_{A} = (\frac {(b+c-a)^2 sin A}{8r})( \frac{1}{cos(B-C) - sinA} - \frac{1}{cos(B-C)+sin A}) = \frac {(b+c-a)^2 (sin A)^2}{16 r sin B sin C cos B cos C}$, using the fact that $ sin A = cos (B+C)$. Thus, $ 8 (P_{A} Q_{A})(P_{B} Q_{B})(P_{C} Q_{C}) = \frac {(b+c-a)^2 (a+b-c)^2 (a+c-b)^2} {2^9 r^3 (cos A cos B cos C)^2} = \frac {r^3} {8 (sin A sin B sin C)^2}$ (using the fact that $ b+c-a = \frac {2r} {tan A}$. Since $ (sin A)^2 = \frac {1 - cos 2A} {2}$, It suffices to prove now that: $ \frac {1}{(1- cos 2A)(1-cos 2B)(1- cos 2C)} \leq (\frac {R}{r})^3$ Using the Cosine Rule, ( $ cos 2A = \frac {b^2 + c^2 - a^2} {2bc}$) we have, $ 1- cos 2A = \frac {(a-b+c)(a-c+b)} {2bc}$, and using the fact that $ \frac {R}{r} = \frac {2abc}{(a-b+c)(a-c+b)(b-a+c)}$, our inequality is reduced to $ 8 ( \frac {abc}{(a-b+c)(a-c+b)(b-c+a)} )^2 \leq 8 ( \frac {abc}{(a-b+c)(a-c+b)(b-c+a)} )^3$, which is clearly true since $ abc \geq (a-b+c)(a-c+b)(b+c-a)$

Let $\angle AOI = \angle P_AOI = \angle Q_AOI = \alpha$. In the $\triangle OP_A I$, $IP_A= R - OP_A + r$. So by cosine rule $OP_A ^2 + OI^2 - 2 \cdot OI \cdot OP_A \cdot \cos \alpha = (R - OP_A + r)^2$. We know that $OI^2 = R^2 - 2Rr$. So we get (1) $OP_A = \frac{4Rr + r^2 }{2R + 2r - 2 \cdot OI \cdot \cos \alpha } $ In $\triangle OQ_A I$, we have $Q_A I = (R - OQ_A - r)^2$. So by cosine rule $OQ_A ^2 + OI^2 - 2 \cdot OI \cdot OQ_A \cdot \cos \alpha = (R - OQ_A - r)^2$. So we get (2) $OQ_A = \frac{r^2}{ 2R - 2r - 2 \cdot OI \cdot \cos \alpha } $ But from $ \triangle AIO$, we have $R^2 + OI^2 - 2\cdot OI \cdot R \cdot \cos \alpha = AI^2 $ Which gives $- 2 \cdot IO \cdot R \cdot \cos \alpha = AI^2 + 2Rr - 2R^2$. Plugging this in equation (1) and (2) yields $OP_A = \frac{Rr^2 + 4R^2 r}{ 4Rr + AI^2}$ and $OQ_A = \frac{Rr^2}{AI^2 } $. So (3) $P_A Q_A = OP_A - OQ_A = P_A Q_A = 4R^2r \left ( \frac{AI^2 - r^2}{AI^2 (4Rr + AI^2)} \right ) $. Now we use the following relations $AI^2 - r^2 = (s-a)^2$. $AI^2 = \frac{(s-a)bc}{s}$ and $4Rr = \frac{abc}{s} $. So $AI^2 + 4Rr = bc$. Plugging these relations in (3) we get $P_A Q_A = \frac{a\cdot R\cdot (s-a)}{bc}$ So the desired inequality becomes $(b+c-a)(c+a-b)(a+b-c) \le abc $, which is true. Equality holds iff $a=b=c$, i.e when the triangle is equilateral.

Do you remember a famous problem which was also proposed by USA ? Suppose we have a triangle $ ABC $ and a circle tangent to $ AB ~,~ AC $ respectively at $ E ~,~ F $ and the circumcircle internally ( externally ) at $ T $ . Then the incentre ( excentre opposite $A$ ) of $ \Delta ABC $ is the mid-pt of segment $ EF $ . It can be proved by joining $ TE $ and $ TF $ and applying Pascal's theorem on $ A B' C T B C' $ , where $ B' ~,~ C' $ are the mid-pts of arc $ AC ~,~ AB $ , which are also the points lying on $ TF ~,~ TE $ respectively . Let $ R_1 ~,~ R_2 $ be the radius of $ \omega_A ~,~ S_A $ respectively , $ I_A $ the excentre opposite $A$ , $M$ the mid-pt of the chord of $\omega$ which is the polar of $A$ . Now we know that $ \frac{P_A Q_A}{R } = \frac{ R_2 - R_1 }{R} = AM \left( \frac{1}{AI} - \frac{1}{AI_A} \right) $ $ P_A Q_A = 2R~\frac{ (a/2)(s-a) }{bc} = \frac{aR(s-a)}{bc} $ Therefore , the inequality is equivalent to : $ 8(s-a)(s-b)(s-c) = (a+b-c)(a-b+c)(-a+b+c) \leq abc $ which is known as Pados' inequality .

It turns out we can compute $P_AQ_A$ explicitly. Let us invert around $A$ with radius $s-a$ (hence fixing the incircle) and then compose this with a reflection around the angle bisector of $\angle BAC$. We denote the image of the composed map via \[ \bullet \mapsto \bullet^\ast \mapsto \bullet^+. \]We overlay this inversion with the original diagram. Let $P_AQ_A$ meet $\omega_A$ again at $P$ and $S_A$ again at $Q$. Now observe that $\omega_A^\ast$ is a line parallel to $S^\ast$; that is, it is perpendicular to $\overline{PQ}$. Moreover, it is tangent to $\omega^\ast = \omega$. Now upon the reflection, we find that $\omega^+ = \omega^\ast = \omega$, but line $\overline{PQ}$ gets mapped to the altitude from $A$ to $\overline{BC}$, since $\overline{PQ}$ originally contained the circumcenter $O$ (isogonal to the orthocenter). But this means that $\omega_A^\ast$ is none other than the $\overline{BC}$! Hence $P^+$ is actually the foot of the altitude from $A$ onto $\overline{BC}$. By similar work, we find that $Q^+$ is the point on $\overline{AP^+}$ such that $P^+Q^+ = 2r$. [asy][asy] size(8.5cm); pair A = dir(140); pair B = dir(210); pair C = dir (330); filldraw(A--B--C--cycle, invisible, grey); dot("$A$", A, A); dot("$B$", B, B); dot("$C$", C, C); draw(incircle(A,B,C), black+1); pair D = foot(A,B,C); dot("$P^+$", D, dir(-45), brown); draw(A--D, brown); pair I = incenter(A,B,C); pair O = circumcenter(A,B,C); real a = abs(B-C); real b = abs(C-A); real c = abs(A-B); real s = 0.5*(a+b+c); real ha = abs(A-D); real k = ((s-a)**3) / (s*ha*abs(A-I)**2); pair PA = k*b*c/2*dir(O-A)+A; dot("$P_A$", PA, dir(255), blue); draw(CP(PA,A), lightblue); pair P = Drawing("P", 2*PA-A, dir(180)); pair QA = PA + dir(PA-A)*( (s-a)*a**2 / (a*b*c) ); dot("$Q_A$", QA, dir(-80)*1.4, blue); draw(CP(QA,A), lightblue); pair Q = Drawing("Q", 2*QA-A, dir(-70)); draw(unitcircle, grey); draw(A--Q); real r = abs(I-foot(I,B,C)); Drawing("I", I, dir(90)); pair X = I + dir(O-A)*r; pair P1 = foot(X,P,Q); draw(Line(X,P1,2.5), red, Arrows(TeXHead)); dot("$P^\ast$", P1, dir(5), red); draw(CP(A, foot(I,A,B)), dashed+deepgreen); pair Q2 = D+dir(A-D)*2*r; dot("$Q^+$", Q2, dir(145), brown); draw(Q2--(I+r*dir(90)), brown); [/asy][/asy] Now we can compute all the lengths directly. We have that \[ AP_A = \frac{1}{2} AP = \frac{(s-a)^2}{2AP^+} = \frac{1}{2} (s-a)^2 \cdot \frac{1}{h_a} \]and \[ AQ_A = \frac{1}{2} AQ = \frac{(s-a)^2}{2AQ^+} = \frac{1}{2} (s-a)^2 \cdot \frac{1}{h_a-2r} \]where $h_a = \frac{2K}{a}$ is the length of the $A$-altitude, with $K$ the area of $ABC$ as usual. Now it follows that \[ P_AQ_A = \frac{1}{2} (s-a)^2 \left( \frac{2r}{h_a(h_a-2r)} \right). \]This can be simplified, as \[ h_a - 2r = \frac{2K}{a} - \frac{2K}{s} = 2K \cdot \frac{s-a}{as}. \]Hence \[ P_AQ_A = \frac{a^2rs(s-a)}{4K^2} = \frac{a^2(s-a)}{4K}. \]Hence, the problem is just asking us to show that \[ a^2b^2c^2(s-a)(s-b)(s-c) \le 8 (RK)^3. \]Using $abc=4RK$ and $(s-a)(s-b)(s-c) = \frac 1s K^2 = rK$, we find that this becomes \[ 2(s-a)(s-b)(s-c) \le RK \iff 2r \le R \]which follows immediately from $IO^2=R(R-2r)$. Alternatively, one may rewrite this as Schur's Inequality in the form \[ abc \ge (-a+b+c)(a-b+c)(a+b-c). \]

Let $r,r_a,r_i,r_e$ be the radius of the incircle, A-excircle, A-internal and A-external mixtilinear circles respectively. Note that $\omega$ is the ex-mixtilinear wrt $S_A, AB, AC$ and in-mixtilinear wrt $\omega_A, AB, AC$. Dilate $S_A$ to $S$ and $\omega_A$ to $S$ to obtain: $\frac{AP_A}{R}=\frac{r}{r_e}$ , $\frac{AQ_A}{R}=\frac{r}{r_i}$ Subtract the last 2 to get $\frac{P_AQ_A}{R}=\frac{r}{r_i}-\frac{r}{r_e}$ It can be easily shown (from well-known lemmas about mixtilinear circles) that $\frac{r}{r_i}=cos^2\frac{A}{2}$ , $\frac{r_a}{r_e}=cos^2\frac{A}{2}$ and $\frac{r}{r_a}=\frac{s-a}{s}$ Combining the last 4 we get $\frac{P_AQ_A}{R}=\frac{a}{s}cos^2\frac{A}{2}$ So the inequality becomes $\frac{8abc}{s^3}cos^2\frac{A}{2}cos^2\frac{B}{2}cos^2\frac{C}{2} \leq 1$ This is however obvious by multipliing $s^3 \geq \frac{27}{8}abc$ (AM-GM) and $cos^2\frac{A}{2}cos^2\frac{B}{2}cos^2\frac{C}{2} \leq (\frac{\sqrt3}{2})^6$ (Jensen)

PP. Let an acute $\triangle ABC$ with the incircle $ \omega = C(I,r)$ and the circumcircle $ \rho = C(O,R)$. The circles $ c_{1}= C\left(P,r_{1}\right)$ and $ c_{2}= \mathbb C\left(Q,r_{2}\right)$ are tangent internally to $ \rho$ in the same point $ A$. The circle $ w$ is tangent externally to the circle $ c_{1}$ and is tangent internally to the circle $ c_{2}$. Prove that $PQ =\frac{a^{2}(p-a)}{4S}$, where $p$ is the semiperimeter and $ S=[ABC]$ is the area for$ \triangle ABC$. Remark. Prove easily that $\left \{\begin{array}{ccccc} IP = r+r_{1} & ; & IQ = r_{2}-r\\\\ PO = R-r_{1} & ; & PA = r_{1}\\\\ QO = R-r_{2} & ; & QA = r_{2}\\\\ OA = R & ; & PQ = r_{2}-r_{1}\end{array}\right\|$. The relations $\left \{\begin{array}{ccc} IO^{2}= R(R-2r) & ; & IA^{2}=\frac{bc(p-a)}{p}=\frac{4Rr(p-a)}{a}\\\\ IA^{2}-r^{2}= (p-a)^{2} & ; & IA^{2}+4Rr = bc\\\\ p(p-a)+(p-b)(p-c) = bc.& ; & p(p-a)(p-b)(p-c) = S^2\end{array}\right\|$ are well-known. Proof I. Denote $\left\{\begin{array}{ccc} IO & = & m\\\ IA & = & n\end{array}\right\|$. Apply the Stewart's theorem in $ \triangle OIA$ for the cevian-rays $ [IP$ and $ [IQ\ :\ \left\{\begin{array}{ccc} m^{2}r_{1}^{2}+n^{2}(R-r_{1}) & = & R(r+r_{1})^{2}+Rr_{1}(R-r_{1})\\\\ m^{2}r_{2}+n^{2}(R-r_{2}) & = & R(r_{2}-r)^{2}+Rr_{2}(R-r_{2})\end{array}\right\|$ $ \implies$ $ r_{1}(R^{2}+2Rr+n^{2}-m^{2}) = R(n^{2}-r^{2}) = r_{2}(R^{2}-2Rr+n^{2}-m^{2})$ $ \implies$ $ r_{1}(R^{2}+2Rr+bc-4Rr-R^{2}+2Rr) =$ $ R(p-a)^{2}=$ $ r_{2}(R^{2}-2Rr+bc-4Rr-R^{2}+2Rr)$ $ \implies$ $ r_{1}bc = R(p-a)^{2}= r_{2}(bc-4Rr)$ $ \implies$ $ PQ = r_{2}-r_{1}=$ $ \frac{4R^{2}r(p-a)^{2}}{bc(bc-4Rr)}=$ $ \frac{4R^{2}pr(p-a)^{2}}{b^{2}c^{2}(p-a)}=$ $ \frac{4R^{2}pra^{2}(p-a)^{2}}{16R^{2}p^{2}r^{2}(p-a)}$ $ \implies$ $ \boxed{\ PQ =\frac{a^{2}(p-a)}{4S}\ }$. Proof II. Apply the Pythagoras' theorem in the triangles : $ \triangle\ IOP\blacktriangleright$ $ (r+r_{1})^{2}= (R-r_{1})^{2}+(R^{2}-2Rr)-2(R-r_{1})\cdot$ $ OI\cdot\frac{(R^{2}-2Rr)+R^{2}-\frac{4Rr(p-a)}{a}}{2R\cdot IO}\implies$ $ \boxed{r_{1}=\frac{R(p-a)^{2}}{bc}}\ .$ $ \triangle\ IOQ\blacktriangleright$ $ (r_{2}-r)^{2}= (R-r_{2})^{2}+(R^{2}-2Rr)-2(R-r_{2})\cdot$ $ IO\cdot\frac{(R^{2}-2Rr)+R^{2}-\frac{4Rr(p-a)}{a}}{2R\cdot IO}\implies$ $ \boxed{r_{2}=\frac{Rp(p-a)}{bc}}\ .$ Therefore, $ \frac{r_{1}}{p-a}=\frac{r_{2}}{p}=\frac{R(p-a)}{bc}=\frac{PQ}{a}$, i.e. $ \boxed{PQ =\frac{aR(p-a)}{bc}}=\frac{a^{2}(p-a)}{4S}$. Observe that $ \frac{r_{1}}{r}=\frac{r_{2}}{r_{a}}$, where $ r_{a}$ is the $ A$- exinradius of $ \triangle ABC$. Remark. Prove easily that two more interesting relations : $ \boxed{\sum PQ = R+r\ }$ and $ r_{1}= R-\frac{a(4R+r)}{4p}$ , $ \boxed{r_{2}=\frac{ r_{b}+r_{c}}{4}}$.

Let $AP_A=x$ and $AQ_A=y$. Then by Stewart's on $AIO$ with $P_A$ on $O$ we get $x(R-x)\cdot R + (r+x)^2\cdot R = (R^2-2Rr)x+AI^2(R-x)$. Solving, $x=\frac{R(AI^2-r^2)}{4rR+AI^2}$. Note that $AI^2-r^2$ is the power of $A$ wrt the incircle, so $AI^2-r^2=(s-a)^2$. Similarly, note $AI^2=\frac{bc}{s}\cdot (s-a)$. Thus $x=\frac{R(s-a)^2}{4bc(s-a)/s+4rR}$. Similarly, by Stewarts on $AIO$ but with $Q_A$ instead, we get $y(R-y)\cdot R + (r-y)^2\cdot R = (R^2-2Rr)y+AI^2(R-y)$, or $y=\frac{R(s-a)^2}{4bc(s-a)/s}$. Now $y-x= R(s-a)^2 \frac{4rR}{\frac{bc}{s}(s-a)\left[\frac{bc}{s}(s-a)+4rR\right]}=\frac{4rR^2s^2(s-a)}{bc(bc(s-a)+4rRs)}$. Note $4rRs=4R[ABC]=abc$ so $y-x=\frac{4rR^2s(s-a)}{b^2c^2}$. The inequality we want to prove becomes $512r^3R^6s^3(s-a)(s-b)(s-c)\le R^3a^4b^4c^4$. Recall that Schur's gives $8(s-a)(s-b)(s-c)\le abc$, with equality for $a=b=c$. On the other hand, $(4rRs)^3=(abc)^3$. Multiplying, we get $512r^3R^3s^3(s-a)(s-b)(s-c)\le a^4b^4c^4$ as desired. $\square$

Consider an inversion about $A$ with radius $\sqrt{AB\cdot AC}$ followed by a reflection across the angle bisector of $\angle BAC$. This takes the incircle to the $A$-mixtilinear-excircle and the circumcircle of $ABC$ to $BC$. Since $\omega_A$ and $S_A$ are circles that are tangent to the circumcircle of $ABC$, their image is parallel to $BC$. Furthermore, they are tangent to the $A$-mixtilinear-excircle. Finally, $AP_A$ and $AQ_A$ map to the altitudes from $A$ to the tangents to the mixtilinear excircle that are parallel to $BC$. Denote the distance from $A$ to these tangents by $d_1$ and $d_2$. We have that $2P_AQ_A=\frac{bc}{d_1}-\frac{bc}{d_2}=\frac{bc(d_2-d_1)}{d_1d_2}=\frac{2bcR_A}{d_1d_2}$. Let $h_a$ be the length of the $A$-altitude, $r$ be the inradius of $ABC$, and $r_a$ be the $A$-exradius. By similar triangles, $d_1=\frac{h_aR_A}{r_a}$ and $d_2=\frac{h_aR_A}{r}$. Hence, $2P_AQ_A=\frac{2bcrr_a}{h_a^2R_A}$. Multiplying similar expressions for $2P_BQ_B$ and $2P_CQ_C$, it suffices to show that $\frac{8a^2b^2c^2r^3r_ar_br_c}{h_a^2h_b^2h_c^2R_AR_BR_C}\leq R^3$ or $8a^2b^2c^2r^3r_ar_br_c\leq R^3h_a^2h_b^2h_c^2R_AR_BR_C$. Because the $A$-excenter lies on the polar of $A$ with respect to $\Omega_A$, we can get $R_A=\frac{r_a}{\cos^2\frac{A}{2}}$ with similar triangles. Hence, it remains to show that $R^3h_a^2h_b^2h_c^2\geq8r^3a^2b^2c^2\cos^2\frac{A}{2}\cos^2\frac{B}{2}\cos^2\frac{C}{2}$. Multiplying both sides by $a^2b^2c^2$, the inequality becomes $8R^3[ABC]^6\geq r^3a^4b^4c^4\cos^2\frac{A}{2}\cos^2\frac{B}{2}\cos^2\frac{C}{2}$. Since $[ABC]=\frac{abc}{4R}$, we can further reduce this to $[ABC]^3\geq8r^3abc\cos^2\frac{A}{2}\cos^2\frac{B}{2}\cos^2\frac{C}{2}$. Since $[ABC]=rs$, this is equivalent to $(a+b+c)^3\geq 64abc\cos^2\frac{A}{2}\cos^2\frac{B}{2}\cos^2\frac{C}{2}$. By AM-GM, $(a+b+c)^3\geq27abc$ so it suffices to show that $\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}\leq\frac{3\sqrt{3}}{8}$. Note that $f(x)=\cos x$ is concave on $[0,\frac{\pi}{2}]$. By Jensen's inequality, $\cos\frac{A}{2}+\cos\frac{B}{2}+\cos\frac{C}{2}\leq\frac{3\sqrt{3}}{2}$. Hence, we are done by AM-GM. Equality holds iff $a=b=c$ on the first AM-GM, so equality holds iff $ABC$ is equilateral.

Here's a much cleaner solution: The central claim is that $\frac{2P_AQ_A}R=\frac{a(b+c-a)}{bc}$. Let $M$ be the midpoint of arc $BC$ not containing $A$, and let $AD$ meet $\omega_A$ and $S_A$ at $X_A$ and $Y_A$. Let $I$ and $DEF$ be the incenter and intouch triangle of $ABC$, and let $D'$ be the reflection of $D$ across $I$. Let $AI$ meet $D'E,D'F,DE,DF,EF$ at $S,T,U,V,W$. By similar circles, we have that $\frac{2P_AQ_A}R=\frac{2X_AY_A}{AM}$. We claim that $X_A$ and $Y_A$ are the midpoints of $ST$ and $UV$. This follows from the configuration of 2016 ELMO P6. Hence, we have $2X_AY_A=2AY_A-2AX_A=AU+AV-AS-AT=US+VT$. We also have that $US=EW\left(\tan\frac B2+\cot\frac B2\right)=\frac{2EW}{\sin B}=\frac{(b+c-a)\sin\frac A2}{\sin B}$ and similarly $VT=\frac{(b+c-a)\sin\frac A2}{\sin C}$. Thus, \begin{align*}\frac{2X_AY_A}{AM}&=\frac{(b+c-a)\sin\frac A2(\sin B+\sin C)}{2R\cos\left(\frac{B-C}2\right)\sin B\sin C}\\&=\frac{2(b+c-a)\sin\frac A2\sin\left(\frac{B+C}2\right)\cos\left(\frac{B-C}2\right)}{2R\cos\left(\frac{B-C}2\right)\sin B\sin C}\\&=\frac{(b+c-a)\sin A}{2R\sin B\sin C}\\&=\frac{a(b+c-a)}{bc}\end{align*}as desired. To finish, note that the inequality is now equivalent to $(b+c-a)(c+a-b)(a+b-c)\le abc$, which is just Schur's inequality. Equality holds for positive $a,b,c$ iff $a=b=c$, as desired. In fact, the stronger inequality $P_AQ_A+P_BQ_B+P_CQ_C=R+r\le\frac{3R}2$ holds and can be proven similarly.

Please tell me there's a better way to construct $\omega_A$ other than wu2481632 is way too extra wrote: Let M_A be the midpoint of minor arc BC Let D be the A-intouch point similarly define M_B, E By Canada 2007/5 we know that M_AD and M_BE concur at the insimilicenter of the incircle and circumcircle Now take O as the circumcenter, I as the incenter it is well-known that (O, similitude center; I, similitude center) forms a harmonic bundle) so we can find the external similitude center by constructing the circle with diameter OI and constructing the polar of the insimilicenter next, we know that A is an external similitude center of omega_A and the circumcircle by Monge's theorem the three exsimilicenters are collinear and one lies on the incircle so we intersect the line joining A and the exsimilicenter of incircle and circumcircle with the incircle which determines our third similitude center we then draw the tangent from this point to the incirlce and the tangent from A to the circumcircle and draw perpendiculars to both tangents at the points of tangency which helps us find the center of omega_A at which point the rest is trivial

In my defense, this process takes about 2 minutes on geogebra and you explicitly said this was allowed.

[asy][asy] import cse5; import olympiad; pathpen = black; pointpen = black; size(13cm); pen n_purple = rgb(0.7,0.4,1), n_blue = rgb(0,0.6,1), n_green = rgb(0,0.4,0), n_orange = rgb(1,0.4,0.1), n_red = rgb(1,0.2,0.4); pair O = origin, A= dir(115), B=dir(210), C=dir(330), I=incenter(A,B,C), D=foot(I,B,C), E=foot(I,C,A), F=foot(I,A,B), XA=I+abs(I-D)*(A-O), YA=2*I-XA, A1=-A, PA1=foot(XA,A,O), QA1=foot(YA,A,O), K=foot(I,A,O), M=dir(-90); DPA(A--B--C--cycle); D(circumcircle(A,B,C), n_blue); D(incircle(A,B,C), n_red); D(A--A1, n_orange); D(CP(A,E), blue); D((XA+(PA1-XA)*6)--(XA-(PA1-XA)*4)); D((YA+(QA1-YA)*6)--(YA-(QA1-YA)*4)); D(I--K, n_purple); D(A--M, n_orange); D(circumcircle(I,M,A1), n_green); D("A",A,A); D("B",B,B); D("C",C,C); D("O",O,dir(0)); D("I",I,dir(160)); D("D",D,dir(90)); D("E",E,2*dir(90)); D("F",F,2*dir(210)); D("Y_A",XA,XA); D("X_A",YA,dir(I-YA)); D("A'",A1,A1); D("Q_A''",PA1,2*PA1); D("P_A''",QA1,2*QA1); D("K",K,dir(60)); D("M",M,M); [/asy][/asy] Let $P_A'$ be the image of $P_A$ under a homothety of ratio $2$ at $A$ (which lies on $\omega_A$), and let $P_A''$ be the image of $P_A'$ under an inversion at $A$ with radius $s-a$ (which fixes the incircle). Let $Q_A''$ be defined similarly. The image of $\omega_A$ under this inversion is the further line tangent to the incircle perpendicular to $AO$, the image of $S_A$ is the closer line (to $A$). Then $P_A'',Q_A''$ are the intersections of these lines with $AO$. Let $K$ be the foot of $I$ onto $AO$. Then $AQ_A''=AK-r,AP_A''=AK+r$. Then $2P_AQ_A=P_A'Q_A'=AQ_A'-AP_A'=(s-a)^2(\frac{1}{AQ_A''}-\frac{1}{AP_A''})=\frac{2r(s-a)^2}{AK^2-r^2}$. Extend $AI$ to meet $S$ again at $M$. Let $A'$ be the antipode of $A$ on $S$. We have $IKA'M$ is cyclic, so $AK = \frac{AI\cdot AM}{AA'}$. By Ptolemy's theorem we have that $AM\cdot a = BM\cdot b+CM\cdot c = (b+c)\cdot MI$, so $AK=\frac{AI\cdot MI}{2R}\cdot\frac{b+c}{a}=\frac{b+c}{a}\cdot\frac{R^2-OI^2}{2R}=\frac{b+c}{a}\cdot r$. So $AK^2-r^2=r^2\Bigg(\bigg(\frac{b+c}{a}\bigg)^2-1\Bigg)=r^2\Bigg(\frac{(b+c-a)(a+b+c)}{a^2}\Bigg)=r^2\frac{2(s-a)\cdot 2s}{a^2}$. Thus $2P_AQ_A=\frac{2r(s-a)^2}{r^2\frac{2(s-a)\cdot 2s}{a^2}}=\frac{a^2(s-a)}{2rs}$. So \[8P_AQ_A\cdot P_BQ_B\cdot P_CQ_C=\frac{(abc)^2(s-a)(s-b)(s-c)}{8K^3}=\frac{(abc)^2s(s-a)(s-b)(s-c)}{8K^3s}=\frac{(abc)^2}{8Ks}=\frac{abcR}{2s}=\frac{2abcrR}{4K}\]\[=2rR^2=(R^2-OI^2)R=R^3-R\cdot OI^2\le R^3.\]$\square$

A dif. Sol (hopefully). I avoid details of the computations to visualise the beauty of the prob. with geo tastes !!! Take an inversion with radius $(s-a)$ and reflection W. r. t. angle bisector of $BAC$ Say $AP_AQ_A$ intersects with $\omega_{A}$ and $S_A$ at $P$ and $Q$ respectively. Under the map $P$ goes to the feet of $A$ altitude and $Q$ goes to a point on $A$ altitude s. t. the distance between the 2 new points is $2r$ where $r$ is inradius. Now all the distance are nicely comparable since we can find the distances in the inverted diagram. The inequality is transformed into $R-2r$ is nonzero. So equally for an equilateral triangle !

[asy][asy] unitsize(0.4inches); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10.548572501878272, xmax = 3.6512772351615244, ymin = -5.1593989481592795, ymax = 10.993869271224643; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); draw((-5.44,7.72)--(-8.4,0.92)--(0.66,1.1)--cycle, linewidth(2) + zzttqq); /* draw figures */ draw((-5.44,7.72)--(-8.4,0.92), linewidth(2) + zzttqq); draw((-8.4,0.92)--(0.66,1.1), linewidth(2) + zzttqq); draw((0.66,1.1)--(-5.44,7.72), linewidth(2) + zzttqq); draw(circle((-4.710258684274337,3.3907645256058427), 2.3969853849513987), linewidth(2)); draw(circle((-3.9097280729330395,3.0096463376296767), 4.952692580342005), linewidth(2)); draw(shift((-4.309993378603689,3.2002054316177597))*rotate(-25.458276913177837)*xscale(3.6748389826467207)*yscale(3.648001708491176)*unitcircle, linewidth(1.2) + dotted); draw(shift((-4.3099933786036875,3.2002054316177593))*rotate(-25.4582769131963)*xscale(1.2778535976953005)*yscale(1.198492692475698)*unitcircle, linewidth(1.2) + dotted); draw((-3.9097280729330395,3.0096463376296767)--(-5.44,7.72), linewidth(2)); draw(circle((-5.129886704468735,6.76543547357939), 1.0036750924307454), linewidth(2)); draw(circle((-4.36587993673596,4.41373443743517), 3.4763667643832514), linewidth(2)); /* dots and labels */ dot((-5.44,7.72),dotstyle); label("$A$", (-5.379526671675426,7.868399699474079), NE * labelscalefactor); dot((-8.4,0.92),dotstyle); label("$B$", (-8.339706987227638,1.0765138993238157), SW * 4*labelscalefactor); dot((0.66,1.1),dotstyle); label("$C$", (0.7211495116453758,1.2568294515401943), SE * labelscalefactor); dot((-3.9097280729330395,3.0096463376296767),linewidth(4pt) + dotstyle); label("$O$", (-3.8468444778362096,3.1351164537941387), NE * labelscalefactor); dot((-4.710258684274337,3.3907645256058427),linewidth(4pt) + dotstyle); label("$I$", (-4.643238166791881,3.5107738542449276), NE * labelscalefactor); dot((-5.129886704468735,6.76543547357939),linewidth(4pt) + dotstyle); label("$P_{A}$", (-5.063974455296764,6.891690458302028), dir(250) *6* labelscalefactor); dot((-4.36587993673596,4.41373443743517),linewidth(4pt) + dotstyle); label("$Q_{A}$", (-4.312659654395187,4.532561983471074), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let the radius of $\omega_A$ be $\rho_A$. We see that $OP_A=R-\rho_A$ and $IP_A=\rho_A+r$. Thus, \[OP_A+IP_A=R+r.\]Similarly, we have $OQ_A+IQ_A=R-r$. Obviously, we have $P_A,Q_A\in OA$. Let $k=OI=\sqrt{R(R-2r)}$, and let $\theta=\angle IOA$. We will use the law of cosines to find $P_AQ_A$. Let $q=OQ_A$. We have that $IQ_A=R-r-OQ_A$, so \[\sqrt{k^2+q^2-2kq\cos\theta}=(R-r)-q.\]Expanding, we see that \[k^2-2kq\cos\theta=(R-r)^2-2q(R-r),\]so $q(R-r-k\cos\theta)=r^2$, or \[q=\frac{r^2}{2(R-r-k\cos\theta)}.\]Let $p=OP_A$. We have $IP_A=R+r-OP_A$, so \[\sqrt{k^2+p^2-2kp\cos\theta}=(R+r)-q,\]and its easy to see that a similar calculation now gives \[p=\frac{r(r+4R)}{2(R+r-k\cos\theta)}.\]Therefore, \begin{align*} P_AQ_A &= p-q \\ &=\frac{r}{2}\left(\frac{r+4R}{R+r-k\cos\theta}-\frac{r}{R-r-k\cos\theta}\right) \\ &=\frac{r}{2}\left(\frac{((R-k\cos\theta)-r)(r+4R)-((R-k\cos\theta)+r)r}{(R+r-k\cos\theta)(R-r-k\cos\theta)}\right) \\ &=\frac{r}{2}\left(\frac{4R(R-k\cos\theta)-r(r+4R)-r^2}{(R+r-k\cos\theta)(R-r-k\cos\theta)}\right) \\ &=r\left(\frac{2R(R-k\cos\theta-r)-r^2}{(R+r-k\cos\theta)(R-r-k\cos\theta)}\right). \end{align*}Yet we also have \[k^2+R^2-2kR\cos\theta=AI^2\implies 2kR\cos\theta=2R^2-2rR-AI^2\implies R-k\cos\theta=r+\frac{AI^2}{2R}.\]Thus, \begin{align*} P_AQ_A &= r\left(\frac{AI^2-r^2}{\left(2r+\frac{AI^2}{2R}\right)\frac{AI^2}{2R}}\right) \\ &=4R^2r\left(\frac{1-(r/AI)^2}{4Rr+AI^2}\right) \\ &= R\left(\frac{\cos^2\frac{A}{2}}{1+\frac{AI^2}{4Rr}}\right). \end{align*}Note that $\frac{abc}{4R}=rs$, so $4Rr=\frac{abc}{s}$. Also, \[AI=\frac{s-a}{\cos\frac{A}{2}}=(s-a)\sqrt{\frac{bc}{s(s-a)}}=\sqrt{bc}\sqrt{\frac{s-a}{s}},\]so \[\frac{AI^2}{4Rr}=\frac{s-a}{a}.\]Thus, \[P_AQ_A=R\frac{\frac{s(s-a)}{bc}}{\frac{s}{a}}=\frac{Ra(s-a)}{bc}.\]We now have \[8\prod_{\mathrm{cyc}}\frac{Ra(s-a)}{bc}=8R^3\frac{(s-a)(s-b)(s-c)}{abc}=R^3\frac{xyz}{\frac{x+y}{2}\frac{y+z}{2}\frac{z+x}{2}},\]where $x=s-a$, $y=s-b$, $z=s-c$ are all positive. By AM-GM, \[\frac{xyz}{\frac{x+y}{2}\frac{y+z}{2}\frac{z+x}{2}}\le 1\]with equality if and only if $x=y=z$, or that $ABC$ is equilateral, so \[8\prod_{\mathrm{cyc}}P_AQ_A\le R^3\]with equality if and only if $ABC$ is equilateral. $\blacksquare$

Solved with Aryan-23. [asy][asy] import cse5; import olympiad; import geometry; pointpen = black; pathpen = black; real p=0.216,q=0.716; pair A=dir(100),B=dir(213),C=dir(342),O=circumcenter(A,B,C),P=p*O+(1-p)*A,Q=q*O+(1-q)*A,I=incenter(A,B,C),M=extension(A,I,O,WP(B--C)),IA=rotate(180,M)*I,BB=extension(IA,rotate(90,IA)*A,A,B),CC=extension(IA,rotate(90,IA)*A,A,C),D=extension(CC,rotate(90,CC)*A,BB,rotate(90,BB)*A); D(D("A",A,A)--D("B",B,B)--D("C",C,C)--cycle); D(circumcircle(A,B,C),blue); D(incircle(A,B,C),red); D(CP(P,A),orange); D(CP(Q,A),deepgreen); D(CP(D,BB),purple); DPA(B--BB^^C--CC); D("P_A",P,S); D("Q_A",Q,S); [/asy][/asy] Consider a $\sqrt{bc}$ inversion. It is well-known that the incircle goes to the $A$-mixtillinear excircle. Now, as $\omega_A$ and $S_A$ pass through $A$, they get inverted to parallel lines tangent to the said excircle. Now, we have that the centers $P_A'$ and $Q_A'$ are the reflections of $A$ in $S_A'$ and $\omega_A'$. Thus, we get \[P_AQ_A=AQ_A-AP_A=\dfrac{bc}{AQ_A'}-\dfrac{bc}{AP_A'}\]However, if $d_1$ and $d_2$ are the distances from $A$ to $S_A'$ and $\omega_A'$, respectively, then we get that \[P_AQ_A=\dfrac 12\left(\dfrac{bc}{d_1}-\dfrac{bc}{d_2}\right)=\dfrac 12\left(\dfrac{bc(d_1-d_2)}{d_1d_2}\right)\]Now, we get that as $\omega_A'\parallel S_A$, we must have that $d_1-d_2=2R_a$, where $R_a$ is the radius of the said $A$-mixtillinear excircle. We note the well-known property that the tangents from $A$ to the mixtillinear excircle are on the perpendicular line from $I_A$ (the $A$-excircle) to $AI$, so thus $I_A$ is the inverse of $A$ with respect to the $A$-excircle. In particular, we have that \[R_a=\dfrac{r_a}{\cos^2\frac 12 A}\]By similar triangles, we must have that as $BC$ and $B'C'$ (where $B'=\omega_A'\cap AB$ and similarly with $C$) are antiparallel, so we must have that \[d_1=\dfrac{R_a\cdot h_a}{r_a}\]by comparing the excircles. Similarly, we get \[d_2=\dfrac{R_a\cdot h_a}{r}\]by comparing incircles. Finally, we also note \[h_a=\dfrac{bc}{2R}\]so thus we get \[P_AQ_A=\dfrac{bcR_arr_a}{R_a^2h_a^2}=\dfrac{4Rr\cos^2\frac 12 A}{bc}\]However, the well-known formula $\cos^2\frac 12A=\frac{s(s-a)}{bc}$, we get \[P_AQ_A=\dfrac{4Rrs(s-a)}{b^2c^2}=\dfrac{4R\Delta(s-a)}{b^2c^2}\]Multiplying the symmetric versions, we get \[8P_AQ_A\cdot P_BQ_B\cdot P_CQ_C=\dfrac{512R^3\Delta^3(s-a)(s-b)(s-c)}{a^4b^4c^4}\]Using the well-known formula \[\Delta=\dfrac{abc}{4R}\]we have \[8P_AQ_A\cdot P_BQ_B\cdot P_CQ_C=\dfrac{8R^3(s-a)(s-b)(s-c)}{abc}\]However, Schur's Inequality guarantees that \[8(s-a)(s-b)(s-c)\leq abc\]so thus we have \[8P_AQ_A\cdot P_BQ_B\cdot P_CQ_C\leq R^3\]completing the proof.

We use the machinery of inversion to determine $P_AQ_A$ etc. Let the intouch triangle be $DEF$, define $a=BC,b=CA,c=AB,s=(a+b+c)/2$. Let the image of $P_A,Q_A$ under transformation of inversion about the circle with radius $AE$ and center $A$ and then reflection over the angle bisector of $\angle BAC$, which we call $\tau$, be $P_A',Q_A'$. Clearly $\tau$ fixes $\omega$. Let the intersection of $AI$ and $\omega$ closer to $A$ be $Q$ and the other be $P$. Then $P_A'$ is the reflection of $A$ over the line $BC$ and $Q_A'$ is the reflection of $A$ over the line parallel to $BC$ through the antipode of $D$ wrt $\omega$, which we subsequently call $\ell$. Let $r$ be the inradius. Then the distance from $A$ to $BC$ is $\frac{2sr}{a}$ so \[2P_AQ_A=\frac{(s-a)^2}{2(s-a)r/a}-\frac{(s-a)^2}{2sr/a} = \frac{a(s-a)}{2r}\cdot (1-(s-a)/s)=\frac{a^2(s-a)}{2rs}.\]Similar results hold for $2P_BQ_B,2P_CQ_C$. It is well-known that $R=\frac{abc}{4[ABC]}$ where $[ABC]$ is the area of $\triangle ABC$: for one proof, use the Law of Sines and the identity $[ABC]=\frac{1}{2}ab\sin C$. Thus it suffices to establish \[\frac{a^2b^2c^2(s-a)(s-b)(s-c)}{(2[ABC])^3}\le \frac{a^3b^3c^3}{(4[ABC])^3}.\]Let $s-a=x,s-b=y,s-c=z$ for positive $x,y,z$, then the result amounts to showing $8xyz\le (x+y)(y+z)(z+x)$ which is immediate by AM-GM on $x+y,y+z,z+x$. Moreover, equality holds exactly when $x=y=z\implies a=b=c$, or when $\triangle ABC$ is equilateral, as desired.

Denote $I,O,x,y,r,s$ as the incenter and circumcenter of $ABC$, radiuses of $\omega_A, S_A,\omega,$ and the semiperimeter of $ABC$, respectively. We first find $P_AQ_A$, from which $P_BQ_B$ and $P_CQ_C$ follow symmetrically. In terms of $x,y,$ $P_AQ_A$ is equivalent to $x-y$. Now by Stewarts Theorem on $AIO$ and cevian $IP_A,$ $$AI^2 (R-x) + OI^2 x = x(R-x)R+(x+r)^2 R$$$$\Rightarrow AI^2(R-x)+R(R-2r)x = x(R-x)R+(x+r)^2R$$$$\Rightarrow AI^2R-x(AI^2-R^2+2Rr) = Rx^2+2Rrx+Rr^2+R^2x-Rx^2$$$$\Rightarrow x(4Rr+AI^2)=R(AI^2-r^2)=R(s-a)^2$$$$\Rightarrow x = \frac{R(s-a)^2}{4Rr+AI^2}$$To solve for $y$ now, we again use Stewarts, this time on $AIQ_A$ and cevian $IP_A:$ $$AI^2(y-x)+(y-r)^2x=(x+r)^2y+x(y-x)y$$$$\Rightarrow AI^2(y-x)-2rxy+r^2x=2rxy+r^2y$$$$\Rightarrow (s-a)^2(y-x)+r^2(y-x)-2rxy+r^2x=2rxy+r^2y$$$$\Rightarrow (s-a)^2(y-x)=4rxy$$$$\Rightarrow y = \frac{x(s-a)^2}{(s-a)^2-4rx}$$So, $$P_AQ_A = y-x = \frac{x(s-a)^2}{(s-a)^2-4rx} - x = \frac{4rx^2}{(s-a)^2-4rx} = \frac{\frac{4r(R^2(s-a)^4)}{4Rr+AI^2}}{(s-a)^2 - \frac{4Rr(s-a)^2}{4Rr+AI^2}}$$$$= \frac{4R^2r(s-a)^2}{(1-\frac{4Rr}{4Rr+AI^2})(4Rr+AI^2)^2} = \frac{4R^2r(s-a)^2}{AI^2(4Rr+AI^2)}$$Hence, it suffices to show $$8\prod_{\text{cyc}} \frac{4R^2r(s-a)^2}{AI^2(4Rr+AI^2)} \leq R^3$$$$\Rightarrow 512R^3r^3(s-a)^2(s-b)^2(s-c)^2 \leq \prod_{\text{cyc}}(AI^2(4Rr+AI^2))$$Using $K=rs$, the $LHS$ simplifies to $512R^3r^7s^2$. Also, $$ \sqrt{\frac{(s-b)(s-c}{bc}} = \sin{(\frac{a}{2})} = \frac{r}{AI} \Rightarrow AI^2 = \frac{r^2bc}{(s-b)(s-c)}$$$$\Rightarrow AI^2+4Rr = \frac{r^2bc}{(s-b)(s-c)} + 4Rr = r(\frac{rbc(s-a)}{(s-a)(s-b)(s-c)} + \frac{abc}{rs})$$$$ = r(\frac{rbc(s-a)+rabc}{r^2s}) = (\frac{bcs}{s}) = bc$$So, the inequality reduces to $$512R^3r^7s^2 \leq \prod_{\text{cyc}} (\frac{r^2bc}{(s-b)(s-c)} bc) = \frac{r^6 a^4b^4c^4}{(s-a)^2(s-b)^2(s-c)^2}$$$$\Rightarrow 512R^3r^7s^2 \leq \frac{r^6 (4Rrs)^4}{r^4s^2}$$$$\Rightarrow 512R^3r^{11}s^4 \leq r^6 (256R^4r^4s^4)$$$$\Rightarrow 2r\leq R$$which is just Euler's Inequality, so we are done $\blacksquare$

Here's a very computational solution with $\sqrt{bc}$ inversion and inversion distance formula (different from post #37). The solution may appear long but the advantage is that there is no tricky step and if we do no silly mistake, then we are just done. Let $r$ be the inradius of $\triangle ABC$ and $2 \alpha,2 \beta, 2 \gamma$ denote the measure of angles $\angle CAB,\angle ABC,BCA$, respectively. Claim (Key Claim): $\displaystyle P_AQ_A = \frac{R \cdot \sin \alpha \cos^2 \alpha}{\cos \beta \cos \gamma}$ Proof: Let $\Gamma$ be the $A$-mixtilinear excircle with center $T$ and radius $R_A$. Let $X,Y$ be the projection of $T$ onto lines $AB,AC$, respectively. It is well known that the midpoint of segment $XY$ is $I_A$, the $A$-excenter of $\triangle ABC$. Thus, $$\sin 2 \alpha \cdot AT_A = XY = 2AI_A \cdot \tan \alpha ~ \implies ~ \boxed{AT_A = \frac{AI_A}{\cos^2 \alpha}}$$Let $\Phi$ be the $\sqrt{bc}$ inversion (followed by reflection in angle bisector, of course). Denote by $Z^*$ the image of a point $Z$ under $\Phi$. Then, $\Phi$ swaps $\omega \longleftrightarrow \Gamma$ and $S \longleftrightarrow \overline{BC}$. $\Phi$ maps $\omega_A,S_A$ to some lines $\ell_1,\ell_2$ (respectively) both parallel to $\overline{BC}$ and tangent to $\Gamma$ such that $\ell_2$ is closer to $A$ than $\ell_1$. $P_A^*,Q_A*$ are reflection of $A$ in lines $\ell_1,\ell_2$, respectively. $\displaystyle P_AQ_A = P_A^* Q_A^* \cdot \frac{bc}{AP_A^* \cdot A Q_A^*}$. Let $\ell_3$ be the line passing through $T$ parallel to $\overline{BC}$ and $\theta = \beta - \gamma$. Now $d(\ell_2,\ell_3) = d(\ell_3,\ell_1) = R_A$ and $d(A,\ell_3) = AT_A \cdot \cos \theta$. Also, $R_A = AT_A \cdot \sin \alpha$. Thus, \begin{align*} P_A^* Q_A^* = 4R_A ~,~ AP_A^* = 2(AT_A \cdot \cos \theta + R_A) ~,~ AQ_A^* = 2(AT_A \cdot \cos \theta - R_A) \\ \implies P_AQ_A = P_A^* Q_A^* \cdot \frac{bc}{AP_A^* \cdot A Q_A^*} = \frac{\sin \alpha}{\cos^2 \theta - \sin^2 \alpha} \cdot \frac{bc}{AT_A} = \frac{r \cdot \cos^2 \alpha}{\cos^2 \theta - \sin^2 \alpha} \end{align*}But we also have $$\cos^2 \theta - \sin^2 \alpha = \Big(\sin(90^\circ + \beta - \gamma) - \sin \alpha \Big)\Big( \sin (90^\circ + \beta - \gamma) + \alpha \Big)$$Now using Product-Sum Identities (in ch-5 of EGMO) we obtain \begin{align*} \sin(90^\circ + \beta - \gamma) - \sin \alpha = 2 \cdot \sin \left( \frac{(90^\circ + \beta - \gamma) - \alpha}{2} \right) \cdot \cos \left( \frac{(90^\circ + \beta - \gamma) + \alpha}{2} \right) = 2 \sin \beta \sin \gamma \\ \sin (90^\circ + \beta - \gamma) + \sin \alpha = 2 \cdot \sin \left( \frac{(90^\circ + \beta - \gamma) + \alpha}{2} \right) \cdot \cos \left( \frac{(90^\circ + \beta - \gamma) - \alpha}{2} \right) = 2 \cos \gamma \cos \beta \end{align*}Hence $$\cos^2 \theta - \sin^2 \alpha = 4 \cos \beta \cos \gamma \sin \beta \sin \gamma $$So we finally have that $$P_AQ_A = \frac{r \cdot \cos^2 \alpha}{\cos^2 \theta - \sin^2 \alpha} = \frac{r \cdot \cos^2 \alpha}{4 \cos \beta \cos \gamma \sin \beta \sin \gamma }$$Lastly, it is well know that $$r = 4R \cdot \sin \alpha \sin \beta \sin \gamma$$Hence, $$P_AQ_A = \frac{r \cdot \cos^2 \alpha}{4 \cos \beta \cos \gamma \sin \beta \sin \gamma } = \frac{R \cdot \sin \alpha \cos^2 \alpha}{\cos \beta \cos \gamma}$$This proves our claim. $\square$ So using our claim, we obtain that the given inequality is equivalent to the following $$\sin \alpha \sin \beta \sin \gamma \le \frac{1}{8} \qquad{(1)}$$The most simple way to finish from here is take $\log$ on both sides and using Jensen's inequality. But we also present a nice synthetic type proof: We know that $$\sin \alpha \sin \beta \sin \gamma = \frac{r}{4R}$$So $(1)$ is equivalent to $$2r \le R$$Which is well known to be true (and we also know that equality holds if and only if circumcenter and incenter of $\triangle ABC$ coincide, which can happen iff and only if $\triangle ABC$ is equilateral). This completes the proof of the problem. $\blacksquare$

Let $H$ be foot from $A$ to $\overline{BC}$. Let $K,L$ be feet from $I$ to $\overline{AO},\overline{AH}$, respectively. Let $R,r,r_1,r_2$ be the circumradius, inradius, radius of $\omega_A$ and radius of $S_A$, respectively. [asy][asy]import olympiad;import geometry; size(10cm);defaultpen(fontsize(10pt)); pair A,B,C,I,O,D,E,H,K,L; A=dir(120);B=dir(205);C=dir(335);O=circumcenter(A,B,C);I=incenter(A,B,C); real R=abs(A-O);real r=abs(I-foot(I,B,C));real k=(2*(abs(A-foot(A,B,C))-r)*R-r^2-2*R*r)/(2(abs(A-foot(A,B,C)))); real l=(2*(abs(A-foot(A,B,C))-r)*R-r^2-2*R*r)/(2*(abs(A-foot(A,B,C))-2r)); D=(A*(R-k)+O*k)/R;E=(A*(R-l)+O*l)/R;H=foot(A,B,C);K=foot(I,O,A);L=foot(I,A,H); draw(A--B--C--cycle);draw(circumcircle(A,B,C));draw(incircle(A,B,C)); draw(circle(D,abs(D-A)));draw(circle(E,abs(E-A))); draw(E--intersectionpoints(line(E,I),incircle(A,B,C))[1]);draw(L--I--foot(I,B,C));draw(A--O);draw(D--I);draw(A--H);draw(I--K); draw(rightanglemark(I,K,O,1));draw(rightanglemark(I,L,A,1));draw(rightanglemark(C,H,A,1));draw(rightanglemark(C,foot(I,B,C),I,1)); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$I$",I,dir(I)); dot("$O$",O,dir(O)); dot("$P_A$",D,dir(D)); dot("$Q_A$",E,dir(E)); dot("$H$",H,dir(H)); dot("$K$",K,dir(K)); dot("$L$",L,dir(L)); [/asy][/asy]By Law of Cosines on $\triangle OIP_A$, \begin{align*} IP_A^2&=OI^2+OP_A^2-2OP_A\cdot OI\cdot \cos{\angle AOI}\implies \\ (r_1+r)^2&=R(R-2r)+(R-r_1)^2-2(R-r_1)(R-AK)\implies \\ r_1&=\frac{2R\cdot AK-r^2-2Rr}{2(AK+r)}. \end{align*}Similarly, by Law of Cosines on $\triangle OIQ_A$, \begin{align*} r_2=\frac{2R\cdot AK-r^2-2Rr}{2(AK-r)}. \end{align*}Hence, \begin{align*} r_2-r_1=\frac{2R\cdot AK-r^2-2Rr}{2(AK-r)}-\frac{2R\cdot AK-r^2-2Rr}{2(AK+r)}=\frac{(2R\cdot AK-r^2-2Rr)r}{AK^2-r^2}. \end{align*}Moreover, since $AK=AL=AH-r=\frac{bc}{2R}-r$, we have \begin{align*} P_AQ_A=r_2-r_1=\frac{(bc-r^2-4Rr)\cdot 4R^2r}{b^2c^2-4bcRr}. \end{align*}Similarly, we find $P_BQ_B$ and $P_CQ_C$, hence we would like to show that \begin{align*} a^2b^2c^2(bc-4Rr)(ab-4Rr)(ac-4Rr)\geq \\ R^3r^3\cdot 2^9\cdot (bc-r^2-4Rr)(ab-r^2-4Rr)(ac-r^2-4Rr). \end{align*}Now, using $Rr=\frac{abc}{2(a+b+c)}$ and $r^2=\frac{(s-a)(s-b)(s-c)}{s}$, we would like to show the following, \begin{align*} abc(s-a)(s-b)(s-c)\geq\\ 2^3 \prod_{cyc} \left(bc-\frac{(s-a)(s-b)(s-c)-abc}{s}\right). \end{align*}Now, for simplicity, we do a sub $x=s-a$, $y=s-b$ and $z=s-c$ and we would like to show that \begin{align*} (x+y+z)^3(x+y)(y+z)(z+x)xyz\geq\\ 2^3\prod_{cyc}\left((x+y+z)(x+y)(y+z)-xyz-(x+y)(y+z)(z+x)\right), \end{align*}which is actually equivalent to \begin{align*} (x+y)(y+z)(z+x)\geq 2^3 xyz, \end{align*}which is true by AM-GM. In fact, the equality holds iff $x=y=z\Longleftrightarrow a=b=c$.

Thanks to Tafi_ak for sharing with me this identity which essentially solves the inequality part which I wasn't being able to do,\[\sin\left(\dfrac{A}{2}\right)\sin\left(\dfrac{B}{2}\right)\sin\left(\dfrac{C}{2}\right)=\dfrac{r}{4R}.\] We already have some exposition on this configuration from ELMO 2016 P6 part (b). Now we perform a $\sqrt{bc}$ Inversion to get that the image of $\odot(S_A)$ which is a line is tangent to the mixtilinear incircle of $\triangle ABC$ apart from being tangent to the image of $\odot(I)$. This helps us to characterize the image of $\odot(I)$ properly, that is, we can now say that the image of $\odot(I)$ is the excircle of the triangle $\triangle AB'C'$ that has it's incircle as the $A-$mixtilinear circle of $\triangle ABC$. Also note that as all the circles are tangent to $\odot(ABC)$ at $A$, we can say that $\overline{A-P_A-Q_A-O}$ are collinear. Now note that the centers of the circles $\odot(\omega_A)$ and $\odot(S_A)$ are just the midpoints of the intersection of $AO$ with $\odot(\omega_A)$ and $\odot(S_A)$ respectively. So their images under this $\sqrt{bc}$ Inversion are going to be the reflections of $A$ over the intersections of the $A-$altitude with the images of $\odot(\omega_A)$ and $\odot(S_A)$ respectively. Let $\ell_1$ and $\ell_2$ be the images of $\odot(S_A)$ and $\odot(\omega_A)$ respectively. Now take the foot from the incenter and do a bunch of sine law span in those triangles to get the radius of the mixtilinear incircle as $=\dfrac{r}{\cos^2\left(\dfrac{A}{2}\right)}$. Now let the intersection of $\ell_1$ and $\ell_2$ with the $A-$altitude be $X$ and $Y$ respectively. Also, let $H_A$ denote the foot of the $A-$altitude. Now taking a homothety at $A$ that sends $\triangle ABC\mapsto\triangle AB'C'$, we get that $\dfrac{AH_A}{AX}=\dfrac{r}{\dfrac{r}{\cos^2\left(\dfrac{A}{2}\right)}}=\cos^2\left(\dfrac{A}{2}\right)$ as their incircles also get mapped. This gives that $AX=\dfrac{AH_A}{\cos^2\left(\dfrac{A}{2}\right)}$. Now we do some more trig spam to get that the radius of the excircle of $\triangle ABC$ is $=s\tan\left(\dfrac{A}{2}\right)$. Now again considering our previous homothety, we get that $\dfrac{s\tan\left(\dfrac{A}{2}\right)}{KL}=\cos^2\left(\dfrac{A}{2}\right)\implies KL=\dfrac{s\sin\left(\dfrac{A}{2}\right)}{\cos^3\left(\dfrac{A}{2}\right)}$. This homothety again finally gives us that $\dfrac{AX}{AY}=\dfrac{\dfrac{r}{\cos^2\left(\dfrac{A}{2}\right)}}{KL}=\dfrac{\dfrac{r}{\cos^2\left(\dfrac{A}{2}\right)}}{\dfrac{s\sin\left(\dfrac{A}{2}\right)}{\cos^3\left(\dfrac{A}{2}\right)}}=\dfrac{r}{s\tan\left(\dfrac{A}{2}\right)}\implies AY=\dfrac{AX\cdot s\tan\left(\dfrac{A}{2}\right)}{r}=\dfrac{AH_A\cdot s\sin\left(\dfrac{A}{2}\right)}{\cos^3\left(\dfrac{A}{2}\right)\cdot r}$. Now we finally use our Inversion-Distance Formula to get $P_AQ_A=\dfrac{1}{2}\cdot\dfrac{bc}{AX\cdot AY}\cdot XY=\dfrac{AB\cdot AC}{AX\cdot AY}\cdot KL=\dfrac{AB\cdot AC\cdot \cos^2\left(\dfrac{A}{2}\right)\cdot r}{AH_A^2}=\dfrac{r\cdot BC^2}{AB\cdot AC\cdot 4\sin^2\left(\dfrac{A}{2}\right)}$. Now literally multiply all the cycles of our current equality to get that what we want to prove is equivalent to proving $R^3\cdot 8\sin^2\left(\dfrac{A}{2}\right)\sin^2\left(\dfrac{B}{2}\right)\sin^2\left(\dfrac{C}{2}\right)\ge r^3$. Now thanks to Tafi_ak , I got this identity from him ,\[\sin\left(\dfrac{A}{2}\right)\sin\left(\dfrac{B}{2}\right)\sin\left(\dfrac{C}{2}\right)=\dfrac{r}{4R},\]and also we have our Euler's Inequality that says $R\ge 2r$ where the inequality holds iff $O\equiv I$ which is when we have an equilateral triangle. Now we just follow through to finish off the rest of the problem. \begin{align*} R^3\cdot 8\sin^2\left(\dfrac{A}{2}\right)\sin^2\left(\dfrac{B}{2}\right)\sin^2\left(\dfrac{C}{2}\right)&=R^3\cdot 8\left(\dfrac{r}{4R}\right)^2\\ &=\dfrac{R\cdot r^2}{2}\\ &\ge r^3 ,\end{align*}which is what we wanted and we are done .

Consider an inversion with radius $s-a$ about $A$ and swap about the $\angle A$-bisector. Define $P$ to be the $A$-antipode in $\omega_A$ and $Q$ similarly, and let $K, L$ be the images of $P, Q$ (typing stars is hard). The conditions mean that $K$ is the foot of the $A$-altitude, and $L$ is the intersection of the $A$-altitude with the tangent to $\omega$ parallel to $\overline{BC}$. Thus, as $AP \cdot AK = (s-a)^2$, we have $AP = \frac{(s-a)^2}{h_A}$ and $AQ = \frac{(s-a)^2}{h_A-2r}$. The rest is computation: this implies $$P_AQ_A = \frac{(s-a)^2 r}{h_A(h_A-2r)} = \frac{(s-a)^2 \cdot \frac Ks}{\frac{2K}a \cdot \left(\frac{2K}a-\frac{2K}s\right)} = \frac{a^2(s-a)}{4K}.$$So then \begin{align*} \prod P_AQ_A &\leq R^3 \\ \iff \frac{a^2b^2c^2 \prod (s-a)}{64K^3} &\leq R^3 \\ \iff a^2b^2c^2 \prod (s-a) &\leq (4KR)^3 \\ \iff \prod (s-a) &\leq abc. \end{align*}This is just Schur, thus done!

One of the worst problems ever.. Invert with radius s-a followed by a reflection over AI (with $a,h_a$ and cyclic variants the length and altitude length usual definition). Let D be the foot of altitude from A, E,F the antipodes of A wrt $\omega_a,S_a$, and s,r the semiperimeter and inradius. S goes to a line parallel to BC (since B',C' lie on AC,AB and satisfy $AC'B'=ABC$), while the incircle is orthogonal and hence preserved so $\omega_a$ that's tangent to the preimage must be a parallel line to the after image and also tangent to $\omega$, so it's BC; in particular, E inverts to D since the inverse lies on BC and satisfies $CAE=CAO=BAD$. Similarly, $S_a$ inverts to the other line l parallel to BC tangent to $\omega$, so F inverts to $AD\cap l$. The problem is bash from here: $AE = \frac{(s-a)^2}{h_a}$ and $AF = \frac{(s-a)^2}{h_a-2r}$ implies \[P_aQ_a=\frac{AF}2-\frac{AE}2= \frac{(s-a)^2 r}{h_A(h_A-2r)}=\frac{(s-a)^2\frac{[ABC]}s}{\frac{2[ABC]}a\left(\frac{2[ABC]}a-\frac{2[ABC]}s\right)}= \frac{a^2(s-a)}{4[ABC]}\]\[\implies8\prod P_AQ_A\le R^3\iff8a^2b^2c^2\prod(s-a)\le(4[ABC]R)^3\iff8\prod(s-a)\le abc\]\(\stackrel{a=x+y,b=y+z,c=z+x}{\iff}8xyz\le(x+y)(y+z)(z+x),\) which is evident by, say, AM-GM over all terms after expanding. $\textbf{Remark.}$ Defining the new points is motivated by the fact that 1. segment connecting A-antipode is diameter, altitude are well known lengths, and because this is inversion unit (but also the circles' tangency suggests that), we look for radii, upon s-a looks fine to work with.

Splendid topic

Consider the inversion at $A$ orthogonal to the incircle followed by a reflection over the angle bisector. Then $P_A,Q_A$ are sent to the reflections of $A$ over $BC$ and the other line parallel to $BC$ tangent to the incircle. Then, by Inversion Distance formula we get that $P_AQ_A=\frac{r(s-a)^2}{h_a(h_a-2r)},$ where $r$ is the inradius, $s$ is the semiperimeter and $h_a$ is the length of the altitude from $A$ to $BC.$ Then, letting $K$ be the area of $ABC$ we get that $h_a=\frac{2K}{a}.$ We proceed similarly for $B,C.$ We then get the following by using Heron's and the inradius and circumradius formula: \begin{align*} 8P_AQ_A \cdot P_BQ_B \cdot P_CQ_C \le R^3 &\iff \frac{8r^3(s-a)^2(s-b)^2(s-c)^2}{\frac{8K^3}{abc}(\frac{2K}{a}-2r)(\frac{2K}{b}-2r)(\frac{2K}{c}-2r)} \le R^3 \\ &\iff \frac{r^3 \frac{K^4}{s^2}}{\frac{K^2}{4R} \cdot 8K^3 \cdot (\frac{1}{a}-\frac{1}{s})(\frac{1}{b}-\frac{1}{s})(\frac{1}{c}-\frac{1}{s})} \le R^3 \\ &\iff \frac{r^3 R}{2Ks^2 \cdot \frac{(s-a)(s-b)(s-c)}{abcs^3}} \le R^3 \\ &\iff \frac{r^3}{2K \cdot \frac{K^2}{4KRs^2}} \le R^2 \\ &\iff \frac{2r^3s^2}{K^2} \le R \\ &\iff 2r \le R, \end{align*}which is well known, e.g. by constructing the nine-point circle.

Let $P_{2A}$ and $Q_{2A}$ denote the reflections of $A$ over $P_A$ and $Q_A$ respectively, which both lie on their respective circles as well as line $AO.$ The problem then reduces to showing that $$\prod_{cyc}P_{2A}Q_{2A}\leq R^3.$$ The main claim is as follows: \begin{claim} $$P_{2A}Q_{2A}=\frac{a^2(-a+b+c)}{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}.$$\end{claim} Perform $\sqrt{bc}$ inversion, and denote image points with $'$. The image of the incircle is the circle tangent to $AB',AC',(AB'C')$. Thus, $Q_{2A}'$ is the intersection of the top parallel tangent to $BC$ and $\omega'$ and the line $AO'$, which is the altitude, and $P_{2A}'$ is the same intersection but with the bottom parallel tangent. Let $T_B'$ and $T_C'$ be the tangents of $\omega'$ to $AB'$ and $AC'$ Note that $$AT_B'=AT_C'=\frac{bc}{s-a}=\frac{2bc}{b+c-a}.$$ If we take a homothety sending $\omega'$ to the excircle, then $Q_{2A}'$ gets sent to the foot from $A$ to $B'C'$, which has length say $L$. Thus, $$AQ_{2A}'=L\cdot\frac{\frac{bc}{s-a}}{s}=\frac{4bc}{(a+b+c)(-a+b+c)}\cdot L.$$Thus, $$AQ_{2A}=\frac{(a+b+c)(-a+b+c)}{4L}$$since the radius of inversion is $\sqrt{bc}$. Now, if we take a homothety sending $\omega$ to the excircle, $P_{2A}'$ gets sent to the foot from $A$ to $B'C'$. Thus, $$AP_{2A}'=L\cdot\frac{\frac{bc}{s-a}}{s-a}=\frac{4Lbc}{(b+c-a)^2}.$$Thus, $$AP_{2A}=\frac{(b+c-a)^2}{4L}.$$This means that $$P_{2A}Q_{2A}=AQ_{2A}-AP_{2A}$$$$=\frac{2a(b+c-a)}{4L}=\frac{a(-a+b+c)}{2L}.$$However, of course we have $$L=\frac{2[ABC]}{a}=\frac{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{2a}.$$Therefore, plugging back in we have $$P_{2A}Q_{2A}=\frac{a^2(-a+b+c)}{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}.$$Thus, we wish to show that $$\prod_{cyc}P_{2A}Q_{2A}\leq R^3$$$$\frac{a^2b^2c^2(-a+b+c)(a-b+c)(a+b-c)}{((a+b+c)(a+b-c)(a-b+c)(-a+b+c))^{3/2}}\leq R^3$$ $$\frac{a^2b^2c^2}{(a+b+c)\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}\leq R^3$$ $$\frac{a^2b^2c^2}{(a+b+c)4A}\leq R^3.$$Substituting $R=\frac{abc}{4A}$ we have $$\frac{a^2b^2c^2}{(a+b+c)4A}\leq \frac{a^3b^3c^3}{64A^3}$$$$\frac{1}{a+b+c}\leq \frac{abc}{16A^2}$$$$16A^2\leq (a+b+c)(abc)$$$$(a+b+c)(a+b-c)(a-b+c)(-a+b+c)\leq (a+b+c)abc$$$$(a+b-c)(a-b+c)(-a+b+c)\leq abc.$$The left hand side expands out to $$(\sum_{sym}a^2b)-(a^3+b^3+c^3)-2abc,$$so it suffices to show that $$a^3+b^3+c^3+3abc\geq \sum_{sym}a^2b,$$which is just Schur's inequality. Since $a,b,c$ are nonzero, equality occurs if and only if $a=b=c.$

Overlay at $A$ with radius $s-a$. Then The incircle remains fixed, while the other three circle get mapped to parallel lines. $AP_AQ_AO$ is mapped to the altitude from $A$ to $BC$. $\omega_A$ is mapped to a line tangent to the incircle and perpendicular to the altitude, which is $BC$. $S_A$ is simiarily mapped to a line parallel to $BC$ tangent to the incircle that is not $BC$. $P_A$ maps to the reflection of $A$ over $\omega_A^*$, and $Q_A$ maps to the reflection of $A$ over $S_A^*$. Consequently, the inversion distance formula says \[P_AQ_A = P_A^*Q_A^* \cdot \frac{(s-a)^2}{AP_A^* \cdot AQ_A^*} = \frac{(s-a)^2 \cdot r}{h_a(h_a-2r)} = \frac{a^2 \cdot (s-a)}{4rs}.\] Thus the LHS can be rewritten as \[8 \cdot \prod_{\text{cyc}} \frac{a^2 \cdot (s-a)}{4[ABC]} = \frac{(abc)^2 \cdot s(s-a)(s-b)(s-c)}{8 s [ABC]^3} = \frac{(4Rrs) \cdot (4R \cdot [ABC]) \cdot [ABC]^2}{8 s [ABC]^3} = 2R^2r,\] which is less than or equal to $R^3$ if and only if $R \ge 2r$, which is true and has equality when $\triangle ABC$ is equilateral. $\blacksquare$

what a bizarre problem Invert at $A$ fixing $\omega$ and then reflect over the $\angle A$-bisector. This has radius $s-a$ and sends $S_A$ and $\omega_A$ to $\overline{BC}$ and the other tangent to $\omega$ parallel to $\overline{BC}$. Moreover, $2P_AQ_A$ equals the distance from $A$ to the $A$-antipode of $S_A$, minus the distance from $A$ to the $A$-antipode of $\omega_A$, and these points get sent to the feet of the altitudes from $A$ to the lines that the circles are mapped to. Hence $$2P_AQ_A=(s-a)^2\left(\frac{1}{d(A,\overline{BC})-2r}-\frac{1}{d(A,\overline{BC})}\right)=(s-a)^2\frac{2r}{h_A(h_A-2r)}.$$It turns out that this actually equals something nice, but we will not realize this and instead turn our brains off. Letting $K=[ABC]$, this implies \begin{align*} 8(P_AQ_A)(P_BQ_B)(P_CQ_C)&=\frac{8r^3(s-a)^2(s-b)^2(s-c)^2}{h_Ah_Bh_C(h_A-2r)(h_B-2r)(h_C-2r)}\\ &=\frac{8r^3K^4}{s^2h_Ah_Bh_C(h_A-2r)(h_B-2r)(h_C-2r)}\\ &=\frac{8r^3K^4}{s^2\frac{8K^3}{abc}(\frac{2K}{a}-2r)(\frac{2K}{b}-2r)(\frac{2K}{c}-2r)}\\ &=\frac{r^3Ka^2b^2c^2}{s^2(2K-2ar)(2K-2br)(2K-2cr)}\\ &=\frac{Ka^2b^2c^2}{s^2(2s-2a)(2s-2b)(2s-2c)}\\ &=\frac{Ka^2b^2c^2}{8sK^2}\\ &=\frac{a^2b^2c^2}{16K^2}\cdot \frac{2K}{s}\\ &=R^2(2r). \end{align*}By Euler's Theorem, $R \geq 2r$ with equality iff $\triangle ABC$ is equilateral, finishing the problem. $\blacksquare$

Lengthy trig bash

$ $ Note that the equation is homogeneous. Without loss of generality (WLOG), assume $R = 2$. We prove that, \[ P_A Q_A \cdot P_B Q_B \cdot P_C Q_C \leq 1 \]with equality if and only if $\triangle ABC$ is equilateral. WLOG, assume that $\angle B > \angle C$. Let $r$ be the inradius and define $x = AP_A$. For $S_A$ to be externally tangent to $\omega$ and internally tangent to $S$, this occurs if and only if $P_A \in AO$ and $P_A I = x + r$. We know \[ \angle IAO = \frac{\angle A}{2} - \angle CAO = \frac{\angle A}{2} - 90^\circ + \angle B = \frac{\angle B}{2} - \frac{\angle C}{2}. \] Using the law of cosines, \[ (x + r)^2 = x^2 + AI^2 - 2x \cdot AI \cdot \cos\left(\frac{B}{2} - \frac{C}{2}\right), \]which simplifies to: \[ x = \frac{AI^2 - r^2}{2(r + AI \cos\left(\frac{B}{2} - \frac{C}{2}\right))}. \] Substituting values: \[ x = \frac{64 \sin^2\frac{B}{2} \sin^2\frac{C}{2} - 64 \sin^2\frac{A}{2} \sin^2\frac{B}{2} \sin^2\frac{C}{2}}{2(8 \sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2} + 8 \sin\frac{B}{2} \sin\frac{C}{2} \cos\left(\frac{B}{2} - \frac{C}{2}\right))}. \] Simplifying further: \[ x = \frac{4 \sin^2\frac{B}{2} \sin^2\frac{C}{2} (1 - \sin^2\frac{A}{2})}{\sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2} + \sin\frac{B}{2} \sin\frac{C}{2} \cos\left(\frac{B}{2} - \frac{C}{2}\right)}, \]\[ x = \frac{4 \sin\frac{B}{2} \sin\frac{C}{2} \cos^2\frac{A}{2}}{\sin\frac{A}{2} + \cos\left(\frac{B}{2} - \frac{C}{2}\right)}, \]\[ x = \frac{4 \sin\frac{B}{2} \sin\frac{C}{2} \cos^2\frac{A}{2}}{\sin\frac{A}{2} + \sin\left(\frac{A}{2} + C\right)}, \]\[ x = \frac{4 \sin\frac{B}{2} \sin\frac{C}{2} \cos^2\frac{A}{2}}{2 \sin\left(\frac{A}{2} + \frac{C}{2}\right) \cos\frac{C}{2}}, \]\[ x = 2 \tan\frac{B}{2} \tan\frac{C}{2} \cos^2\frac{A}{2}. \] On the other hand, let $y$ be the length of $AQ_A$. We know that $Q_A$ lies on $AO$ and $IQ_A = y - r$. Using similar steps: \[ (y - r)^2 = y^2 + AI^2 - 2y \cdot AI \cdot \cos\left(\frac{B}{2} - \frac{C}{2}\right), \]which simplifies to: \[ y = \frac{AI^2 - r^2}{2(AI \cos\left(\frac{B}{2} - \frac{C}{2}\right) - r)}. \] After substitution: \[ y = 2 \cos^2\frac{A}{2}. \] Therefore: \[ P_A Q_A = 2 \cos^2\frac{A}{2} (1 - \tan\frac{B}{2} \tan\frac{C}{2}), \]\[ P_A Q_A = 2 \cos^2\frac{A}{2} \frac{\tan\frac{B}{2} + \tan\frac{C}{2}}{\tan\left(\frac{B}{2} + \frac{C}{2}\right)}, \]\[ P_A Q_A = \sin A (\tan\frac{B}{2} + \tan\frac{C}{2}), \]\[ P_A Q_A = \sin A \frac{\sin\frac{B}{2} \cos\frac{C}{2} + \cos\frac{B}{2} \sin\frac{C}{2}}{\cos\frac{B}{2} \cos\frac{C}{2}}, \]\[ P_A Q_A = \sin A \frac{\cos\frac{A}{2}}{\cos\frac{B}{2} \cos\frac{C}{2}}. \] The desired product is: \[ \frac{\sin A \sin B \sin C}{\cos\frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2}} = 8 \sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2}. \] Since $\sin(x)$ has a second derivative $-\sin(x)$, it is concave for $x \in (0, \frac{\pi}{2})$. By Jensen's inequality: \[ 8 \sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2} \leq 8 \sin^3(30^\circ) = 1. \] Equality holds if and only if $A = B = C = 60^\circ$ as desired.