<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] Put $a = \sin x, b = \cos x \implies a^2+b^2=1\implies \dfrac{a^2}{b}+\dfrac{b^2}{a} > \dfrac{a^2}{1} + \dfrac{b^2}{1} = a^2+b^2 = 1$ . In fact this is a different inequality, but still looks nice though. To return to the question above, observe that $\dfrac{a}{b}+\dfrac{b}{a} \ge 2$ by AM-GM inequality. Thus $(a^2+b^2)\left(\dfrac{a}{b}+\dfrac{b}{a}\right) \ge 2\implies \dfrac{a^3}{b} + \dfrac{b^3}{a} + 2ab \ge 2\implies \dfrac{a^3}{b}+\dfrac{b^3}{a} \ge 2 -2ab  \ge 2 - (a^2+b^2) = 2-1 = 1$ by AM-GM inequality again.
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