Saudi Arabia Mathematical Competitions

Since $a,b\in \left(0,\frac{\pi }{2}\right)$ we have $\sin a,\sin b,\cos a,\cos b>0$. Then $$\begin{array}{rl}\frac{{\sin }^{3}a}{\sin b}+\frac{{\cos }^{3}a}{\cos b}& =\frac{{\sin }^{4}a}{\sin a\sin b}+\frac{{\cos }^{4}a}{\cos a\cos b}\\ & =\frac{({\sin }^{2}a{)}^2}{\sin a\sin b}+\frac{({\cos }^{2}a{)}^2}{\cos a\cos b}\\ & \ge \frac{({\sin }^{2}a+{\cos }^{2}a{)}^2}{\cos a\cos b+\sin a\sin b}=\frac{1}{\cos (a-b)}\end{array}$$ 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 $a=b$.