<s>[INST] 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) ?
 [/INST] Too long for a comment. 
Put $X=4R^2\cos A\cos B \cos C$. 
We have to show that 
$$-\frac 1{\sqrt{2}}>\cos (HIO)=\frac{HI^2+OI^2-OH^2}{2\cdot HI\cdot OI}.$$
Since it is assmed that $2\cdot HI\cdot OI>0$, we have to show that 
$$OH^2- HI^2-OI^2>\sqrt{2}\cdot HI\cdot OI$$
$$2Rr-2r^2-X>\sqrt{2(2r^2-X)(R^2-2Rr)}$$
Since $HI^2>0$ and $OI^2>0$, $2r^2>X$ and $R>2r$, so $2Rr-2r^2-X>0$. Thus we have to show that  
$$(2Rr-2r^2-X)^2>2(2r^2-X)(R^2-2Rr)$$
$$X^2+2XR^2+4Xr^2+4r^4>8RrX$$
The respective quadratic equation for $X$ has a discriminant $D=(R-2r)^2R(R-4r)$ and roots $X_1=4Rr-2r^2-R^2-\sqrt{D}$ and $X_2=4Rr-2r^2-R^2+\sqrt{D}$. So if $R<4r$ then $D<0$ and the inequality is proved. 
Otherwise I tried to evaluate $X$ in terms of $R$ and $r$. According to exercises at p.23 of “Topics in Inequalities - Theorems and Techniques” by Hojoo Lee (version February 25, 2006), we have $X=s^2-(2R+r)^2$. Moreover, in W. J. Blundon (see Problem E1935, Amer. Math. Monthly 73 (1966), 1122) found the best possible inequalities of the form $A(R,r)\le s^2\le B(R, r)$, where $A(x, y)$ and $B(x, y)$ are real quadratic forms $\alpha x^2+\beta xy+\gamma y^2$, namely $$16Rr-5r^2\le s^2\le 4R^2+4Rr+3r^2.$$
This implies $$12Rr-4R^2-6r^2\le X\le 2r^2.$$
Unfortunately, this doesn’t help because  
$$12Rr-4R^2-6r^2\le X_1\le X_2\le 2r^2.$$
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