Let $S$ be the set of all triangles $ABC$ for which
$ 5\left( \dfrac{1}{AP} + \dfrac{1}{BQ} + \dfrac{1}{CR} \right) - \dfrac{3}{\min\{ AP, BQ, CR \}} = \dfrac{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. How do I Prove that all triangles in $S$ are isosceles and similar to one another.
After taking $s-a=x$, $s-b=y$, $s-c=z$ and assuming WLWOG that $x \le y \le z$ after much of grueling work I got an equation
$0 = 4y^2z^2 + 25x^2y^2 + 25x^2z^2 - 16xy^2z - 16xyz^2 + 14x^2yz$
How do I solve this equation to get the required result?
The question you wrote is wrong, it should be $$5\left( \dfrac{1}{AP} + \dfrac{1}{BQ} + \dfrac{1}{CR} \right) - \dfrac{3}{\min\{ AP, BQ, CR \}} = \dfrac{6}{r},\\ and\ not\ \left( \dfrac{1}{AP} + \dfrac{1}{BQ} + \dfrac{1}{CR} \right) - \dfrac{3}{\min\{ AP, BQ, CR \}} = \dfrac{6}{r}, $$ This is your answer
I didn't write the answer cause it would take much time , sorry for that.