Here's a nice proof by contradiction.
Let the incenter $I$ lie on the Euler line of $ABC$.
It is known that orthocenter $H$ and circumcenter $O$ are isogonal conjugates, i.e. $AI$ is the bisector of angle $HAO$.
So, (if the point $A$ does not lie on Euler line) $HA/AO=HI/IO$ (angle bisector theorem).
Also $HB/BO=HC/CO=HI/IO=HA/AO$.
And we know that all points $X$, such that $YX/ZX=const$, lie on a circle with center on the line $YZ$ (Appolonius circle)
So, $A, B, C$ and $I$ lies on the same circle, and that cannot be true.
We have assumed that all of points $A, B, C$ don't lie on the Euler line, so, one of them lies on Euler line and that means $ABC$ is isosceles.
If Coordinate Geometry is allowed, WLOG we can assume $\displaystyle B(0,0),A(a,0), C(0,a)$
So, the equation of $AC$ will be $\displaystyle\frac xa+\frac ya=1\iff x+y-a=0$
So, the centroid will be $\displaystyle C\left(\frac a3,\frac a3\right)$
Now, if $I(p,q)$ then we have the perpendicular distance of $I$ from $AB,BC,CA$ will be same
$\displaystyle\implies |p|=|q|=\frac{|p+q-a|}{\sqrt{1^2+2^2}} $
If $\displaystyle p,q>0,p=q=\frac{2p-a}{\sqrt2}\implies (2-\sqrt2)p=a\iff p=\frac a{2+\sqrt2}$
Now, $IC=|\sqrt{(p-a)^2+(q-a)^2}|=\sqrt2|p-a|$
Can you take it from here?
Best Answer
This is simple to tackle through exact barycentric coordinates: $$ G=\frac{A+B+C}{3},\qquad I=\frac{aA+bB+cC}{a+b+c} \tag{1}$$ give: $$ 3(a+b+c)(G-I) = (b+c-2a) A + (a+c-2b) B + (a+b-2c) C \tag{2}$$ and by assuming that the origin is in the circumcenter $O$ we get:
$$ 9(a+b+c)^2\left\|G-I\right\|^2 = 6R^2(a^2+b^2+c^2-ab-ac-bc)+\ldots \tag{3}$$ that boils down to a quite complicated expression.