<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] Correct me if wrong:
Another option.
Let $x \in (0,π/2)$.
$f(x):=$
$2 \dfrac{\sin^4 x}{2\sin x \cos x} +2 \dfrac{\cos^4 x}{2 \sin x \cos} =$
$\dfrac{2}{\sin 2x}×$
$((\sin^2x+\cos^2x)^2 -(1/2)\sin^2 2x)$
$=\dfrac{2}{\sin 2x}(1-(1/2)\sin^2 2x)$;
Let $y:= \sin 2x$, then $0 <y \le 1,$
and consider $g(y): = 2(1/y - (1/2)y)$.
Need to find the minimum of $g(y)$.
$g'(y) = 2(-1/y^2 -1/2) <0.$
$g$ is strictly decreasing 
$\min_{0<y \le1} g(y)= 1.$
The minimum of the given expression is at $y=\sin 2x =1$,
i.e. at  $x=π/4$.
</s>