Problem 1157

8. Let's assume
$$\alpha=\angle B A C \leqslant \beta=\angle C B A \leqslant \gamma=\angle A C B \text {. }$$

Let the lengths of sides $BC$, $CA$, and $AB$ of $\triangle ABC$ be $a$, $b$, and $c$ respectively. Let $D$ and $E$ be points on side $BC$ such that $B_{1} D \parallel AB$ and $B_{1} E$ bisects $\angle B B_{1} C$. Since $\angle B_{1} D B = 180^{\circ} - \beta$ is an obtuse angle,
$$\Rightarrow B B_{1} > B_{1} D$$
$$\Rightarrow \frac{B E}{E C} = \frac{B B_{1}}{B_{1} C} > \frac{D B_{1}}{B_{1} C} = \frac{A B}{A C} = \frac{B A_{1}}{A_{1} C}$$
$$\Rightarrow B E > B A_{1}$$
$$\Rightarrow \frac{1}{2} \angle B B_{1} C = \angle B B_{1} E > \angle B B_{1} A_{1}.$$
Similarly, $\frac{1}{2} \angle B B_{1} A > \angle B B_{1} C_{1}$.
Thus, $\angle A_{1} B_{1} C_{1} = \angle B B_{1} A_{1} + \angle B B_{1} C_{1}$
$$< \frac{1}{2} \left( \angle B B_{1} C + \angle B B_{1} A \right) = 90^{\circ},$$

which means $\angle A_{1} B_{1} C_{1}$ is an acute angle.
By symmetry, $\triangle A_{1} B_{1} C_{1}$ is an acute triangle.
Let $B B_{1}$ intersect $A_{1} C_{1}$ at point $F$.
Since $\alpha \leqslant \gamma$, we have $a \leqslant c$.
This implies $B A_{1} = \frac{c a}{b + c} \leqslant \frac{a c}{a + b} = B C_{1}$.
Thus, $\angle B C_{1} A_{1} \leqslant \angle B A_{1} C_{1}$.
Since $\angle B_{1} F C_{1} = \angle B F A_{1} \leqslant 90^{\circ}$,
$$\Rightarrow \text{points } H \text{ and } C_{1} \text{ are on the same side of } B B_{1}.$$
Therefore, point $H$ is inside $\triangle B B_{1} C_{1}$.
Similarly, since $\alpha \leqslant \beta \leqslant \gamma$, point $H$ is also inside $\triangle C C_{1} B_{1}$ and $\triangle A A_{1} C_{1}$.
Since $\alpha \leqslant \beta \leqslant \gamma$, we have $\alpha \leqslant 60^{\circ} \leqslant \gamma$.
Thus, $\angle B I C \leqslant 120^{\circ} \leqslant \angle A I B$.
If $\angle A I C \geqslant 120^{\circ}$, let the rotation transformation with center $A$ and angle $-60^{\circ}$ map points $B$, $I$, and $H$ to points $B^{\prime}$, $I^{\prime}$, and $H^{\prime}$, as shown in Figure 7. Then points $B^{\prime}$ and $C$ are on opposite sides of $AB$.
Since $\triangle A I^{\prime} I$ is an equilateral triangle, we have
$$\begin{array}{l} A I + B I + C I = I I^{\prime} + B^{\prime} I^{\prime} + I C \\ = B^{\prime} I^{\prime} + I^{\prime} I + I C. \end{array}$$

Similarly,
$$\begin{array}{c} A H + B H + C H = H H^{\prime} + B^{\prime} H^{\prime} + H C \\ = B^{\prime} H^{\prime} + H^{\prime} H + H C. \\ \text{Since } \angle A I I^{\prime} = \angle A I^{\prime} I = 60^{\circ}, \\ \angle A I^{\prime} B^{\prime} = \angle A I B \geqslant 120^{\circ}, \end{array}$$

and $\angle A I C \geqslant 120^{\circ}$, the quadrilateral $B^{\prime} I^{\prime} I C$ is non-concave and on the same side of $B^{\prime} C$ as point $A$.

Since point $H$ is inside $\triangle A C C_{1}$, i.e., outside the quadrilateral $B^{\prime} I^{\prime} I C$, and $H$ is also inside $\triangle A B I$, point $H^{\prime}$ is inside $\triangle A B^{\prime} I^{\prime}$.
This indicates that point $H^{\prime}$ is outside the quadrilateral $B^{\prime} I^{\prime} I C$.
Thus, the quadrilateral $B^{\prime} I^{\prime} I C$ is inside the quadrilateral $B^{\prime} H^{\prime} H C$.
From equations (1) and (2), we have
$$A H + B H + C H \geqslant A I + B I + C I.$$

If $\angle A I C < 120^{\circ}$, let the rotation transformation with center $C$ and angle $60^{\circ}$ map points $B$, $I$, and $H$ to points $B^{\prime}$, $I^{\prime}$, and $H^{\prime}$. Then points $B^{\prime}$ and $A$ are on opposite sides of $BC$, and similarly to the case where $\angle A I C \geqslant 120^{\circ}$, we get
$$A H + B H + C H \geqslant A I + B I + C I.$$