$\star$ Show that choosing $\alpha = \ln(r/\mu)$ minimizes the right-hand side of inequality (C.47).

This is simple. Let:

$$ f(x) = \exp(mu e^{\alpha} - r) $$

Let's explore it's critical points:

$$ f'(x) = \exp(\mu e^{\alpha} - r)(\mu e^{\alpha} - r) = 0$$

This has a solution for $\alpha = \ln(r/\mu)$. It's easy to see that left of this point $f'(x)$ is negative and right of it it is positive. Thus, it is a local minima.