Problem: Let $S$ be the set of all triangles $ABC$ for which
\[ 5 \left( \frac{1}{AP} + \frac{1}{BQ} + \frac{1}{CR} \right)
  - \frac{3}{\min\{ AP, BQ, CR \}} = \frac{6}{r}, \]
where $r$ is the inradius and $P$, $Q$, $R$ are the points of tangency
of the incircle with sides $AB$, $BC$, $CA$ respectively.
Prove that all triangles in $S$ are isosceles and similar to one another.

Solution: We will prove the inequality
\[ \frac{2}{AP} + \frac{5}{BQ} + \frac{5}{CR} \ge \frac 6r \]
with equality when $AP : BQ : CR = 1 : 4 : 4$.
This implies the problem statement.

Letting $x= AP$, $y = BQ$, $z = CR$, the inequality becomes
\[ \frac2x + \frac5y + \frac5z \ge 6\sqrt{\frac{x+y+z}{xyz}}. \]
Squaring both sides and collecting terms gives
\[
  \frac{4}{x^2} + \frac{25}{y^2} + \frac{25}{z^2} + \frac{14}{yz}
  \ge \frac{16}{xy} + \frac{16}{xz}.
\]
If we replace $x=1/a$, $y=4/b$, $z=4/c$,
then it remains to prove the inequality
\[ 64a^2 + 25(b+c)^2 \ge 64a(b+c) + 36bc \]
where equality holds when $a=b=c$.
This follows by two applications of AM-GM:
\begin{align*}
  16 \left( 4a^2 + (b+c)^2 \right) &\ge 64a(b+c) \\
  9(b+c)^2 &\ge 36bc.
\end{align*}

Again one can tell this is an inequality by counting
degrees of freedom.