IMO Shortlist, 2016

Without loss of generality, assume $\alpha = \angle BAC \leqslant \beta = \angle CBA \leqslant \gamma = \angle ACB$. Denote by $a$, $b$, $c$ the lengths of $BC$, $CA$, $AB$ respectively. We first show that triangle $A_{1}B_{1}C_{1}$ is acute.
Choose points $D$ and $E$ on side $BC$ such that $B_{1}D \parallel AB$ and $B_{1}E$ is the internal angle bisector of $\angle BB_{1}C$. As $\angle B_{1}DB = 180^{\circ} - \beta$ is obtuse, we have $BB_{1} > B_{1}D$. Thus,
$$\frac{BE}{EC} = \frac{BB_{1}}{B_{1}C} > \frac{DB_{1}}{B_{1}C} = \frac{BA}{AC} = \frac{BA_{1}}{A_{1}C}$$
Therefore, $BE > BA_{1}$ and $\frac{1}{2} \angle BB_{1}C = \angle BB_{1}E > \angle BB_{1}A_{1}$. Similarly, $\frac{1}{2} \angle BB_{1}A > \angle BB_{1}C_{1}$. It follows that
$$\angle A_{1}B_{1}C_{1} = \angle BB_{1}A_{1} + \angle BB_{1}C_{1} < \frac{1}{2}(\angle BB_{1}C + \angle BB_{1}A) = 90^{\circ}$$
is acute. By symmetry, triangle $A_{1}B_{1}C_{1}$ is acute.
Let $BB_{1}$ meet $A_{1}C_{1}$ at $F$. From $\alpha \leqslant \gamma$, we get $a \leqslant c$, which implies
$$BA_{1} = \frac{ca}{b + c} \leqslant \frac{ac}{a + b} = BC_{1}$$
and hence $\angle BC_{1}A_{1} \leqslant \angle BA_{1}C_{1}$. As $BF$ is the internal angle bisector of $\angle A_{1}BC_{1}$, this shows $\angle B_{1}FC_{1} = \angle BFA_{1} \leqslant 90^{\circ}$. Hence, $H$ lies on the same side of $BB_{1}$ as $C_{1}$. This shows $H$ lies inside triangle $BB_{1}C_{1}$. Similarly, from $\alpha \leqslant \beta$ and $\beta \leqslant \gamma$, we know that $H$ lies inside triangles $CC_{1}B_{1}$ and $AA_{1}C_{1}$.

As $\alpha \leqslant \beta \leqslant \gamma$, we have $\alpha \leqslant 60^{\circ} \leqslant \gamma$. Then $\angle BIC \leqslant 120^{\circ} \leqslant \angle AIB$. Firstly, suppose $\angle AIC \geqslant 120^{\circ}$.
Rotate points $B$, $I$, $H$ through $60^{\circ}$ about $A$ to $B'$, $I'$, $H'$ so that $B'$ and $C$ lie on different sides of $AB$. Since triangle $AI'I$ is equilateral, we have
$$\begin{equation*} AI + BI + CI = I'I + B'I' + IC = B'I' + I'I + IC \tag{1} \end{equation*}$$
Similarly,
$$\begin{equation*} AH + BH + CH = H'H + B'H' + HC = B'H' + H'H + HC \tag{2} \end{equation*}$$
As $\angle AII' = \angle AI'I = 60^{\circ}$, $\angle AI'B' = \angle AIB \geqslant 120^{\circ}$ and $\angle AIC \geqslant 120^{\circ}$, the quadrilateral $B'I'I C$ is convex and lies on the same side of $B'C$ as $A$.
Next, since $H$ lies inside triangle $ACC_{1}$, $H$ lies outside $B'I'I C$. Also, $H$ lying inside triangle $ABI$ implies $H'$ lies inside triangle $AB'I'$. This shows $H'$ lies outside $B'I'I C$ and hence the convex quadrilateral $B'I'I C$ is contained inside the quadrilateral $B'H'H C$. It follows that the perimeter of $B'I'I C$ cannot exceed the perimeter of $B'H'H C$. From (1) and (2), we conclude that
$$AH + BH + CH \geqslant AI + BI + CI.$$
For the case $\angle AIC < 120^{\circ}$, we can rotate $B$, $I$, $H$ through $60^{\circ}$ about $C$ to $B'$, $I'$, $H'$ so that $B'$ and $A$ lie on different sides of $BC$. The proof is analogous to the previous case and we still get the desired inequality.