Let $s$, $R$ and $r$ be the semi-perimeter, circumradius and inradius of $\triangle A{}BC$. Let ${}I$, ${}O$, $G$ be the incentre, circumcentre and centroid of $\triangle A{}BC$. Then $OG$ is the Euler line and \begin{align*} OI^2&=R^2-2 r R,~IG^2=\frac{5 r^2-16 r R+s^2}{9}, \\ OG^2&=\frac{2 r^2+8 r R+9 R^2-2 s^2}{9} .\\ \end{align*}If $R<\left(1+\sqrt2\right)r$, then $OI^2<r^2$, meaning that $O$ is inside the incircle, which contradicts. So assume $R\ge\left(1+\sqrt2\right)r$. By Heron Formula, \[[OIG]^2=\frac{-r^4-12 r^3 R-48 r^2 R^2-2 r^2 s^2-64 r R^3+20 r R s^2+4 R^2 s^2-s^4}{144}.\]If the distance from $I{}$ to $OG$ is $d$, then \begin{align*} r^2-d^2&=r^2 -\left(\frac{2[OIG]}{OG}\right)^2\\&=\frac{9 r^4+44 r^3 R+84 r^2 R^2-6 r^2 s^2+64 r R^3-20 r R s^2-4 R^2 s^2+s^4}{4 \left(2 r^2+8 r R+9 R^2-2 s^2\right)}. \end{align*}Assume that $\triangle A{}BC$ is acute, we claim that $r^2-d^2>0$, or \begin{align}9 r^4+44 r^3 R+s^2 \left(-6 r^2-20 r R-4 R^2\right)+84 r^2 R^2+64 r R^3+s^4>0.\end{align}Case1. $R\ge4r$. Note that $(1)$ is quadratic in $s^2$, and the axis of symmetry is at $m:=2R^2+10Rr+3r^2$. Since $s\ge2R+r$, \[s^2>m\Leftarrow R^2\ge r^2+3rR\Leftarrow R^2\ge4Rr\ge3rR+r^2,\]which is true. So it suffices to prove $(1)$ with $s=2R+r$, and we get $(1)\iff4r^3(2R+r)>0$, done. Case2. $\left(1+\sqrt2\right)r\le R<4R$. We have \[\text{LHS of }(1)=\left(-3 r^2-10 r R-2 R^2+s^2\right)^2+4 R (4 r-R) \left(R^2-2Rr-r^2\right)>0.\]Therefore, $(1)$ is proven. This means that $OG$ passes through the interior of the incircle, which contradicts. So we're done.