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) ?

This formula calculates the $θ$ angle formed by Orthocenter(H),Incenter(I) and circumcenter(O) of ABC triangle (sides:a,b,c).
$θ(a,b,c)=acot\left [\left [ \frac{(-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2)}{(-a+b+c)(a-b+c)(a+b-c)}-2abc+(-a+b+c)(a-b+c)(a+b-c) \right ] \sqrt{\frac{(-a+b+c)(a-b+c)(a+b-c)}{4(a+b+c)^3(a-b)^2(a-c)^2(b-c)^2}}  \right ] $
$θ(3,10,11)= π -acot(\frac{523}{1344}\sqrt{6})=133.627°$
defining $θ$ as a function in the interval of existence of the ABC triangle to use the Calculus tools
$\begin{array}{} |b-c|<a<b+c & a≠b & a≠c \end{array} $
$Range(θ)=(\frac{ π }{2}, π )$
$θ=\frac{ π }{2}$ in the right triangle, $θ(\sqrt{b^2+c^2-2bc·cos(\frac{ π }{2})},b,c)$ whith $c→∞$ or $c→0$. $θ=π$ for $a=|b-c|$, $a=b$, $a=c$ and $a=b+c$. 
maximum and minimum
$\begin{array}{} \frac{d}{da}θ(a,b,c)=0 & solve&for&a &  \end{array}$ 
$\begin{array}{} r1=2.43938 & r2=10.54162 & r3=12.39471 \end{array}$
$θ(r3,b,c)=132.292°$
$\begin{array}{} θ(a,10,11)=135°=\frac{3}{4}π & a1=1.82913 & a2=3.27918 \end{array}$

$\begin{array}{} \frac{R}{r}<4 & ? \end{array}$

$k=\frac{R}{r}=\frac{2abc}{(-a+b+c)(a-b+c)(a+b-c)}$
