# An inequality in triangle geometry involving triangle OIH

Let triangle ABC be such that the points O,I,H are not collinear (O,I,H being the circumcenter, incenter and orthocenter respectively.) Show that $2R_{OIH} > R_{ABC}$, where $R_{ABC}$ and $R_{OIH}$ are the circumradiuses of triangles ABC, OIH.

I was able to prove it only for obtuse or right triangles. It is well known that $\angle OIH > 90$, thus $\sin OIH < 1$. Next, by the sine law we obtain that $\dfrac{OH}{\sin{OIH}} = 2R_{OIH}$ thus the inequality we have to show becomes $\dfrac{OH}{\sin{OIH}} > R_{ABC}$, or equivalently $\dfrac{OH}{R} > \sin{OIH}$. Since $OH \geq R$ when $ABC$ is obtuse or right, we obtain that $\dfrac{OH}{R} \geq 1 > \sin{OIH}$ and thus the inequality is proven.

How could one go about proving the inequality for acute triangles? I tried many tricks, such as using the formulas $$R = \dfrac{abc}{4S} , S_{OIH} = \dfrac{1}{8r}(a-b)(b-c)(c-a)$$ in order to obtain an expression for $R_{OIH}$; I have also tried to use the Mitrinovici inequality, which in this case says $R_{OIH} \geq \dfrac{2}{3 \sqrt{3}} (OI + IH + OH)$, using which one arrives at the following inequality, after applying the formulas for the three distances: $$14R^2 + 10Rr + 5r^2 - 3p^2 > \dfrac{3\sqrt{3}}{2} R ,$$ where $R,r,p$ are the circumradius, inradius and semiperimeter of ABC. I have not been able to prove this inequality.

Another interesting and possibly useful fact is that $\angle IOH < \dfrac{\pi}{6}.$ However, using this in order the bound the sinus of $\angle IOH$ is not strong enough to prove the inequaity.

Source for almost all inequalities mentioned: https://en.wikipedia.org/wiki/List_of_triangle_inequalities

The formula for the area of triangle OIH can be found in Titu Andreescu's Complex Numbers from A to Z.

By Feuerbach's theorem the incircle is tangent to the nine-point-circle, whose radius equals $\frac{1}{2}R_{ABC}$, since the nine-point-circle is the circumcircle of the median triangle. It follows that the distance of $I$ from the midpoint of $OH$ is exactly $\frac{R}{2}-r$, and the circumradius of $OIH$ is larger than the circumradius of an isosceles triangle with base $OH$ and height $\frac{R}{2}-r$. It is enough to show

$$(R-2r)^2+OH^2 \geq 2R(R-2r)$$ which is equivalent to $$8R^2+4r^2 \geq a^2+b^2+c^2\qquad \Longleftrightarrow\quad OH^2+4r^2\geq R^2$$ just a slight strengthening of $9R^2\geq a^2+b^2+c^2$.