69th Belarusian Mathematical Olympiad

Note that $\angle ABM=\angle ACB$ by the property of the angle between the tangent to the circle at point $B$ and the chord $AB$. The right triangles $AMB$ and $ANC$ are similar since they have equal acute angles, therefore $\frac{AM}{AN}=\frac{AB}{AC}$, whence $AN=\frac{AM\cdot AC}{AB}$.

By analogy, the triangles $ANB$ and $ALC$ are similar and $\frac{AL}{AN}=\frac{AC}{AB}$, whence $AN=\frac{AL\cdot AB}{AC}$. Therefore $$A{N}^2=\frac{AM\cdot AC}{AB}\cdot \frac{AL\cdot AB}{AC}=AM\cdot AL.$$ $$\text{Finally, from the AM GM inequality }AM+AL\ge 2\sqrt{AM\cdot AL}=2\sqrt{A{N}^2}=2AN.$$