Angle formed by orthocenter, incenter and circumcenter of a triangle $>135^\circ$? If $H$ is the orthocenter, $I$ the incenter and $O$ the circumcenter of a triangle , the I need to show that the angle $HIO>135^\circ$ 
With the assumptions of $OI^2=R^2-2Rr$, $OH^2=9R^2-(a^2+b^2+c^2)$, $HI^2=2r^2-4R^2\cos A\cos B \cos C$ and $R^2\cdot8(1+\cos A\cos B\cos C)= a^2+b^2+c^2$ 
Applying the $cosine$ rule  I got to something like $$\cos (HIO)=\frac{2r^2-2Rr+R^2\cdot 4\cos A\cos B\cos C}{2(2r^2-4R^2\cos A\cos B\cos C)(R^2-2Rr)}$$
I need to show that the LHS is between $\left[-\frac{1}{\sqrt{2}},-1\right]$
How t0 proceed?
Is there any simpler way to prove the question (without applying cosine rule) ?

It is not always true that $\angle{HIO}\gt 135^\circ$.
A counterexample is $(a,b,c)=(85,13,88)$.
$\qquad\qquad$
For $(a,b,c)=(85,13,88)$, we have
$$\begin{align}R&=\frac{abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}=\frac{2431}{4\sqrt{186}}
\\\\r&=\frac{1}{2}\sqrt{\frac{(-a+b+c)(a-b+c)(a+b-c)}{a+b+c}}=\frac{40\sqrt{186}}{93}
\\\\OI^2&=R^2-2Rr=\frac{1451307}{992}
\\\\OH^2&=9R^2-(a^2+b^2+c^2)=\frac{2712387}{992}
\\\\HI^2&=2r^2+4R^2-\frac{a^2+b^2+c^2}{2}=\frac{109875}{248}
\\\\\cos(\angle{HIO})&=\frac{HI^2+OI^2-OH^2}{2\cdot HI\cdot OI}=-\frac{13693}{\sqrt{708721585}}\approx -0.5143524
\\\\\angle{HIO}&\approx 120.95^\circ\end{align}$$
