Centroid on the Incircle in Right Triangle
We know that
Since $IG=r,\,$ we derive $a^2+b^2+c^2=6r^2+24Rr,\,$ implying that what we need to prove is
$\displaystyle 648Rr\ge 25(6r^2+24Rr),$
or, equivalently, $48Rr\ge 25\cdot 6r^2,\,$ i.e., $8R\ge 25r,\,$ which is a well-known inequality.