62nd Ukrainian National Mathematical Olympiad

To prove this, we will show that the following inequalities hold: $A<\frac{3}{2}<B$.

$$A=\frac{1}{9}:\sqrt[3]{\frac{1}{2023}}=\frac{\sqrt[3]{2023}}{9}<\frac{13}{9}<\frac{3}{2},B={\log }_{2023}91125>{\log }_{2023}45^3={\log }_{45^2}45^3=\frac{3}{2}.$$