Taiwan IMO Selection Camp

Denote $\angle BAC = \alpha$, $\angle CBA = \beta$, $\angle ACB = \gamma$. Without loss of generality, we may assume $\alpha \le \beta \le \gamma$.
Also denote the three side lengths of triangle $ABC$ as $BC = a$, $CA = b$, $AB = c$.

We first prove: $\triangle A_1B_1C_1$ is also an acute triangle. Take points $D, E$ on side $BC$ satisfying: $B_1D // AB$, and $B_1E$ is the interior angle bisector of $\angle BB_1C$. Since $\angle B_1DB = 180^\circ - \beta$ is obtuse, we know $BB_1 > B_1D$. Thus we have
$$\frac{BE}{EC} = \frac{BB_1}{B_1C} > \frac{DB_1}{B_1C} = \frac{BA}{AC} = \frac{BA_1}{A_1C}.$$
From this we know $BE > BA_1$, and $\frac{1}{2}\angle BB_1C = \angle BB_1E > \angle BB_1A_1$. Similarly we obtain $\frac{1}{2}\angle BB_1A > \angle BB_1C_1$. Therefore
$$\angle A_1B_1C_1 = \angle BB_1A_1 + \angle BB_1C_1 < \frac{1}{2}(\angle BB_1C + \angle BB_1A) = 90^\circ$$
is acute. By symmetry, we conclude that $\triangle A_1B_1C_1$ is an acute triangle.

Returning to the original problem. Let line $BB_1$ meet $A_1C_1$ at point $F$. From $\alpha \le \gamma$, we know $a \le c$, and thus we have
$$BA_1 = \frac{ca}{b+c} \le \frac{ac}{a+b} = BC_1$$
so $\angle BC_1A_1 \le \angle BA_1C_1$. Because $BF$ is the interior angle bisector of $\angle A_1BC_1$, $\angle B_1FC_1 = \angle BFA_1 \le 90^\circ$. Therefore $H$ and $C_1$ lie on the same side of line $BB_1$, so $H$ lies inside triangle $BB_1C_1$. Similarly, since $\alpha \le \beta$ and $\beta \le \gamma$, we know $H$ lies inside triangle $CC_1B_1$, and also inside triangle $AA_1C_1$.

Since $\alpha \le \beta \le \gamma$, we have $\alpha \le 60^\circ \le \gamma$. Hence $\angle BIC \le 120^\circ \le \angle AIB$. We first discuss the case $\angle AIC \ge 120^\circ$.

Rotate each of the points $B, I, H$ by $60^\circ$ about the center $A$, obtaining points $B', I', H'$ respectively, such that $B'$ and $C$ lie on opposite sides of line $AB$. Since $\triangle AI'I$ is an equilateral triangle, we know
$$AI + BI + CI = I'I + B'I' + IC = B'I' + I'I + IC. \quad (1)$$
Similarly we know
$$AH + BH + CH = H'H + B'H' + HC = B'H' + H'H + HC. \quad (2)$$
Since $\angle AII' = \angle AI'I = 60^\circ$, $\angle AI'B' = \angle AIB \ge 120^\circ$, and $\angle AIC \ge 120^\circ$, $B'I'IC$ is a convex quadrilateral, lying on the same side of line $B'C$ as point $A$.

Next, since $H$ lies inside triangle $ACC_1$, $H$ lies outside quadrilateral $B'I'IC$. Also, since $H$ lies inside triangle $ABI$, we get that $H'$ also lies inside triangle $AB'I'$. Therefore $H'$ also lies outside $B'I'IC$. Hence, the quadrilateral $B'I'IC$ lies entirely inside quadrilateral $B'H'HC$. From this we know the perimeter of $B'I'IC$ does not exceed the perimeter of $B'H'HC$. Thus from (1) and (2) we obtain
$$AH + BH + CH \geq AI + BI + CI.$$

When $\angle AIC < 120^\circ$, we may rotate points $B, I, H$ by $60^\circ$ about the center $C$, obtaining points $B', I', H'$ respectively, such that $B'$ and $A$ lie on opposite sides of $BC$. The proof in this case is similar to the case above, and yields the same inequality. This completes the proof.