$S$, $I$, $O$ are circumcenter, incenter and orthocenter then $SO\ge IO \sqrt2$

Question

Let $S$, $I$ and $O$ be the circumcenter, incenter and orthocenter of $\triangle ABC$ then prove that $SO\ge IO \sqrt2$, or equivalently $SO^2\ge 2IO^2$. I was able to derive an expression for $SO^2$ using the cosine law: $$SO^2=R^2(1+4\cos^2A+2\cos A [ \cos 2B+\cos 2C])$$ I was able to get an expression for $IO^2$ as well but it was too long, and seemed unfit for this question. Besides that, I can't think of any useful tricks in this inequality. Can anyone help?

Answer 1

From this, $\displaystyle IO^2=2r^2+4R^2-S_w,$ where $\displaystyle2S_w=a^2+b^2+c^2$

From this, $\displaystyle SO^2=9R^2-2S_w$

So, $\displaystyle SO^2-2IO^2=R^2-4r^2$

But from this, $\displaystyle R\ge 2r$