<s>[INST] 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. 
 [/INST] Your proof is right.
Also, by C-S we obtain:
$$\frac{\sin^3x}{\cos x} +\frac{\cos^3x}{\sin x}=\frac{\sin^4x}{\sin{x}\cos x} +\frac{\cos^4x}{\sin x\cos{x}}\geq\frac{(\sin^2x+\cos^2x)^2}{2\sin{x}\cos{x}}=\frac{1}{\sin2x}\geq1.$$
The equality occurs for $x=\frac{\pi}{4},$ which says that $1$ is a minimal value.
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