guess no one bothered with this one, but I'll give an approach (I'll leave it for those that care to finish... ) Two steps: first show that angleCBH < angleCBI < angleCBO (I lies within angle HBO), this is really easy once you simplify everything. Second, prove that IY (Y being the perpendicular foot of I on AB) is not long enough to reach line segment OH. This is proven by assigning X and Z to be the perpendicular feet of H and O on AB, respectively. Then, showing that (|HX|-|OZ|)*(|YZ|/|XZ|) > |IY|-|OZ|. I give fair warning that this is a nasty and commutative trig thing, it's really not a pleasure doing it. But it works!

Here's a nice way to do it. Since that $O, H$ are isogonal conjugates, $AI, BI, CI$ bisects $\angle HAO, \angle HBO, \angle HCO$ as well. Let $AI, BI, CI$ intersect $OH$ at $K_a, K_b, K_c$, drop perpendiculars $OP_a, OP_b, OP_c$ to $BC, AC, AB$. Since $AB>AC>BC$, $OPc<OP_b<OP_a \Rightarrow CH<BH<AH$ from the well-known fact that $AH=2OP_a$ and similarly for $B, C$. It follows that $\frac{CH}{CO}<\frac{BH}{BO}<\frac{AH}{AO}$, which means \[\frac{HK_c}{K_cO}<\frac{HK_b}{K_bO}<\frac{HK_a}{K_aO}\]Therefore $AI$ intersects segment $OK_b$ and $CI$ intersects segment $HK_b$, therefore they concur inside $\bigtriangleup BOH$, as desired.