Rearrangement inequality and minimal value of $\frac{\sin^3x}{\cos x} +\frac{\cos^3x}{\sin x}$ For $x \in \left(0, \dfrac{\pi}{2}\right)$, is the minimum value of $\dfrac{\sin^3x}{\cos x} +\dfrac{\cos^3x}{\sin x} = 1$? So considering ($\dfrac{1}{\cos x}$, $\dfrac{1}{\sin x}$) and ($\sin^3x$, $\cos^3x$), is it right to use the rearrangement inequality and conclude that $\dfrac{\sin^3x}{\cos x} +\dfrac{\cos^3x}{\sin x}$ is more than or equal to $\dfrac{\sin^3x}{\sin x} +\dfrac{\cos^3x}{\cos x}$ which is equal to $1$? Thanks. 

Yes it is fine, indeed for $x\in(0,\pi/2)$ we have in both cases


*

*$\sin x\ge \cos x \implies \sin^3 x\ge \cos^3 x \implies \frac1{\sin x}\le \frac1{\cos x}$


then
$$\frac{\sin^3 x}{\cos x}+\frac{\cos^3 x}{\sin x}\ge \frac{\sin^3 x}{\sin x}+\frac{\cos^3 x}{\cos x}=1$$


*$\cos x\ge \sin x \implies \cos^3 x\ge \sin^3 x \implies \frac1{\cos x}\le \frac1{\sin x}$


then
$$\frac{\sin^3 x}{\cos x}+\frac{\cos^3 x}{\sin x}\ge \frac{\sin^3 x}{\sin x}+\frac{\cos^3 x}{\cos x}=1$$
and equality holds for $\sin x=\cos x$ that is $x=\pi/4$.
