Question 162
Let $\triangle ABC$ be an acute triangle with angles $\alpha, \beta$, and $\gamma$. Prove that
$$
\frac{\cos \alpha}{\cos (\beta-\gamma)}+\frac{\cos \beta}{\cos (\gamma-\alpha)}+\frac{\cos \gamma}{\cos (\alpha-\beta)} \geq \frac{3}{2}
$$
$$
\frac{\cos \alpha}{\cos (\beta-\gamma)}+\frac{\cos \beta}{\cos (\gamma-\alpha)}+\frac{\cos \gamma}{\cos (\alpha-\beta)} \geq \frac{3}{2}
$$