Question
If there is a triangle $ABC$ where its circumcentre, incentre and centroid are named $O, I, G$ respectively, $AB=c$, $BC=a$, $CA=b$, and the radii of the circumcircle and incircle of the triangle are $R$ and $r$ respectively, prove that
$$R^2-OG^2\ge4(r^2-IG^2)$$
Working
After thinking about this question for a little, as it is from a course, I watched the explanation. It used a theorem (of which I unfortunately forgot its name) where it states that
If $G$ is the centroid of a triangle, and $P$ is any other point inside the triangle not on the circumference, $PA^2+PB^2+PC^2=\frac13(a^2+b^2+c^2)+3PG^2$.
So the course teacher said that we should have
\begin{align} OA^2+OB^2+OC^2&=\frac13(a^2+b^2+c^2)+3OG^2\\ R^2+R^2+R^2&=\frac13(a^2+b^2+c^2)+3OG^2\\ 3R^2-3OG^2&=\frac13(a^2+b^2+c^2)\\ R^2-OG^2&=\frac19(a^2+b^2+c^2) \end{align}
And also that
\begin{align} IA^2+IB^2+IC^2&=\frac13(a^2+b^2+c^2)+3IG^2 \end{align}
We can drop a perpendicular from the incircle of $ABC$ so that $IM\perp AB$ at $M$, so $IA^2=r^2+AM^2$ by Pythagoras, and then $IA^2=r^2+(P-a)^2$, where $P$ is the semi-perimeter of the triangle. By the same logic, $IB^2=r^2+(P-b)^2$ and $IC^2=r^2+(P-c)^2$, so
\begin{align} IA^2+IB^2+IC^2&=\frac13(a^2+b^2+c^2)+3IG^2\\ 3r^2+(P-a)^2+(P-b)^2+(P-c)^2&=\frac13(a^2+b^2+c^2)+3IG^2\\ \end{align} \begin{align} r^2-IG^2&=\frac13\left[\frac13(a^2+b^2+c^2)-(P-a)^2-(P-b)^2-(P-c)^2\right]\\ &=\frac19\left[(a^2+b^2+c^2)-(P-a)^2-(P-b)^2-(P-c)^2\right]\\ &=\frac19\left[a^2+b^2+c^2-P^2-a^2+2aP-P^2-b^2+2bP-P^2-c^2+2cP\right]\\ &=\frac19\left[-3P^2+2aP+2bP+2cP\right]\\ &=-\frac13P^2+\frac19\left[a(a+b+c)+b(a+b+c)+c(a+b+c)\right]\\ &=-\frac13(a+b+c)^2\cdot\frac14+\frac19(a+b+c)^2\\ &=\left(\frac19-\frac1{12}\right)(a+b+c)^2\\ r^2-IG^2&=\frac1{36}(a+b+c)^2\\ 4(r^2-IG^2)&=\frac19(a+b+c)^2 \end{align}
But this is a contradiction, as $a,b,c$ are all positive, so we have proved that
$$R^2-OG^2\le 4(r^2-IG^2)$$
Can anyone please tell me where the course teacher got it wrong? Thank you!