The 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 $AB{H}_1$ and $AC{H}_1$ respectively (Fig.23). Then $AB$ and $AC$ are diameters of these circles, as $\angle A{K}_{2}B=\angle A{K}_{3}C=90^{\circ }$, so $K_2\in {\omega }_1$ and $K_3\in {\omega }_2$. Then $AB\ge K_2{H}_1$ and $AC\ge K_3{H}_1$.

Fig.23

Next, by Ptolemy's theorem $AB\cdot K_2{H}_1=A{K}_2\cdot B{H}_1+B{K}_2\cdot A{H}_1$. This implies $$K_2{H}_1=\frac{A{K}_2\cdot B{H}_1}{AB}+\frac{B{K}_2\cdot A{H}_1}{AB}=\frac{A{H}_2}{AB}\cdot B{H}_1+\frac{B{H}_2}{AB}\cdot A{H}_1=B{H}_1\cos \alpha +h\sin \alpha .$$ Analogously, $K_3{H}_1=C{H}_1\cos \alpha +h\sin \alpha$. Then $$AB+AC\ge K_2{H}_1+K_3{H}_1=(B{H}_1\cos \alpha +h\sin \alpha )+(C{H}_1\cos \alpha +h\sin \alpha )=BC\cos \alpha +2h\sin \alpha ,$$ the proof is finished.

Equality holds if and only if the segments $K_2{H}_1$ and $K_3{H}_1$ are the diameters of circles ${\omega }_1$ and ${\omega }_2$ respectively. In this case $$\angle ABC=90^{\circ }-\angle BA{H}_1=\angle BA{K}_2=\angle BAC=\angle CA{K}_3=90^{\circ }-\angle CA{H}_1=\angle ACB,$$ then the triangle $ABC$ is equilateral. It is easily checked that the equality holds.