My solution: Let $ \text{R}, \text{r} $ be the radius of $ \odot (O), \odot (I) $, respectively . Let $ N $ be the 9-point center of $ \triangle ABC $ and $ T $ be the reflection of $ O $ in $ I $ . Since $ \overline{HT}=2 \cdot \overline{NI}=\text{R}-2 \text{r} $ , so $ \overline{HT} < \sqrt{{\text{R}}^2-2\text{R} \text{r}}=\overline{OI}=\overline{TI} \Longrightarrow \angle HIT<90^{\circ} \Longrightarrow \angle OIH>90^{\circ} $ . Q.E.D

I think the following is stronger: http://artofproblemsolving.com/community/c6h39447p246214

More general, we prove that $\widehat{OIH} \ge 90^{\circ}+\arcsin \sqrt{\tfrac{2r}{R}}.$ Where $r$ and $R$ denote the radii of the incircle and circumcircle of $\triangle ABC.$ Denoting by $\varrho=\cos A \cos B \cos C,$ we get $OI^2=R^2-2Rr,$ $IH^2=2(r^2-2R^2\varrho)$ and $OH^2=R^2(1-8\varrho).$ By cosine law for $\triangle OIH,$ we get then $2 \cdot OI \cdot IH \cdot \cos \widehat{OIH}=OI^2+IH^2-OH^2=$ $= -2(Rr-r^2-2R^2\varrho)=-\left ( \frac{2r}{R} \cdot OI^2+IH^2 \right) \Longrightarrow$ $2 \cdot OI \cdot IH \cdot \sin (\widehat{OIH}-90^{\circ})= \frac{2r}{R} \cdot OI^2+IH^2.$ But $\frac{2r}{R}OI^2+IH^2 \ge 2 \sqrt{ \frac{2r}{R}} \cdot OI \cdot IH$ with equality iff $\sqrt{ \frac{2r}{R}} \cdot OI =IH.$ Hence $\sin (\widehat{OIH}-90^{\circ}) \ge \sqrt{ \frac{2r}{R}} \Longrightarrow \widehat{OIH} \ge 90^{\circ}+\arcsin \sqrt{\tfrac{2r}{R}}.$

Moreover, if $ \triangle ABC $ is an acute triangle we can prove $ \angle OIH>135^{\circ} : $ Let $ \mu =\cos \angle A \cos \angle B \cos \angle C >0. $ We need to prove $$ \cos \angle OIH=\frac{{OI}^2+{HI}^2-{OH}^2}{2 \cdot OI \cdot HI}< \frac{-1}{\sqrt{2}} $$$$ \Longleftrightarrow ({OI}^2+{HI}^2-{OH}^2)^2>2 \cdot {OI}^2 \cdot {HI}^2 $$$$\Longleftrightarrow (\text{R}\text{r}-{\text{r}}^2-2 {\text{R}}^2 \mu)^2>({\text{R}}^2-2\text{R}\text{r})({\text{r}}^2-2 {\text{R}}^2 \mu) $$$$ \Longleftrightarrow ({\text{r}}^2-2 {\text{R}}^2 \mu)^2 + 2 \mu ({\text{R}}^2-2 \text{R} \text{r} )^2 >0 $$(See also here)

Izho problem