Problem: Let $ABC$ be an acute triangle
with $\omega$, $S$, and $R$ being its incircle,
circumcircle, and circumradius, respectively.
Circle $\omega_{A}$ is tangent internally to $S$
at $A$ and tangent externally to $\omega$.
Circle $S_{A}$ is tangent internally to $S$
at $A$ and tangent internally to $\omega$.

Let $P_A$ and $Q_A$ denote
the centers of $\omega_A$ and $S_A$, respectively.
Define points $P_B$, $Q_B$, $P_C$, $Q_C$ analogously.
Prove that
\[ 8 P_A Q_A \cdot P_B Q_B \cdot P_C Q_C \le R^3 \]
with equality if and only if triangle $ABC$ is equilateral.

\end{enumerate}

Solution: 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 $\ol{PQ}$.
Moreover, it is tangent to $\omega^\ast = \omega$.

Now upon the reflection,
we find that $\omega^+ = \omega^\ast = \omega$,
but line $\ol{PQ}$ gets mapped to the altitude from $A$ to $\ol{BC}$,
since $\ol{PQ}$ originally contained the circumcenter $O$
(isogonal to the orthocenter).
But this means that $\omega_A^\ast$ is none other than the $\ol{BC}$!
Hence $P^+$ is actually the foot of the altitude from $A$
onto $\ol{BC}$.

By similar work, we find that $Q^+$
is the point on $\ol{AP^+}$ such that $P^+Q^+ = 2r$.
\begin{center}
\begin{asy}
  size(8.5cm);
  pair A = dir(140);
  pair B = dir(210);
  pair C = dir (330);
  filldraw(A--B--C--cycle, opacity(0.1)+yellow, gray);
  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, gray);
  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);
\end{asy}
\end{center}

Now we can compute all the lengths directly.  We have that
\[ AP_A = \half AP = \frac{(s-a)^2}{2AP^+}
= \half (s-a)^2 \cdot \frac{1}{h_a} \]
and
\[ AQ_A = \half AQ = \frac{(s-a)^2}{2AQ^+}
= \half (s-a)^2 \cdot \frac{1}{h_a-2r} \]
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
\[ P_AQ_A = \half (s-a)^2 \left( \frac{2r}{h_a(h_a-2r)} \right). \]
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
\[ abc \ge (-a+b+c)(a-b+c)(a+b-c). \]