Problems of Ukrainian Authors

If $\alpha \ge 90^\circ$, then $AB + AC > 2h \ge 2h \sin \alpha \ge BC \cdot \cos \alpha + 2h \sin \alpha$, because $\cos \alpha \le 0$, that is, the inequality is strict.

Let now $\alpha < 90^\circ$, denote by $H_1, H_2, H_3$ the feet of altitudes from vertices $A, B, C$ respectively, $K_2$ be the point symmetric to $H_2$ with respect to line $AB$ and analogously $K_3$ be the point symmetric to $H_3$ with respect to line $AC$. Denote by $\omega_1$ and $\omega_2$ the circumscribed circles of the triangles $ABH_1$ and $ACH_1$ respectively (Fig.23). Then $AB$ and $AC$ are diameters of these circles, as $\angle AK_2B = \angle AK_3C = 90^\circ$, so $K_2 \in \omega_1$ and $K_3 \in \omega_2$. Then $AB \ge K_2H_1$ and $AC \ge K_3H_1$.

Fig.23

Next, by Ptolemy's theorem $AB \cdot K_2H_1 = AK_2 \cdot BH_1 + BK_2 \cdot AH_1$. This implies
$$K_2H_1 = \frac{AK_2 \cdot BH_1}{AB} + \frac{BK_2 \cdot AH_1}{AB} = \frac{AH_2}{AB} \cdot BH_1 + \frac{BH_2}{AB} \cdot AH_1 = BH_1 \cos \alpha + h \sin \alpha.$$
Analogously, $K_3H_1 = CH_1 \cos \alpha + h \sin \alpha$. Then
$$AB + AC \geq K_2H_1 + K_3H_1 = (BH_1 \cos \alpha + h \sin \alpha) + (CH_1 \cos \alpha + h \sin \alpha) = BC \cos \alpha + 2h \sin \alpha,$$
the proof is finished.

Equality holds if and only if the segments $K_2H_1$ and $K_3H_1$ are the diameters of circles $\omega_1$ and $\omega_2$ respectively. In this case
$$\angle ABC = 90^\circ - \angle BAH_1 = \angle BAK_2 = \angle BAC = \angle CAK_3 = 90^\circ - \angle CAH_1 = \angle ACB,$$
then the triangle $ABC$ is equilateral. It is easily checked that the equality holds.