Saudi Arabian Mathematical Competitions

Since $a, b \in \left(0, \frac{\pi}{2}\right)$ we have $\sin a, \sin b, \cos a, \cos b > 0$.
Then
$$\begin{aligned} \frac{\sin^{3} a}{\sin b} + \frac{\cos^{3} a}{\cos b} &= \frac{\sin^{4} a}{\sin a \sin b} + \frac{\cos^{4} a}{\cos a \cos b} \\ &= \frac{(\sin^{2} a)^2}{\sin a \sin b} + \frac{(\cos^{2} a)^2}{\cos a \cos b} \\ &\geq \frac{(\sin^{2} a + \cos^{2} a)^2}{\cos a \cos b + \sin a \sin b} = \frac{1}{\cos(a-b)} \end{aligned}$$
where we have used the Cauchy-Schwarz inequality. We have equality if and only if
$$\frac{\sin^{2} a}{\sin a \sin b} = \frac{\cos^{2} a}{\cos a \cos b}$$
that is $\sin a \cos b - \sin b \cos a = 0$, hence $\sin(a-b) = 0$ yielding $a = b$.