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. 
 A: 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 $x=\pi/4$.
A: 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.
A: 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.
A: 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  $x=π/4$.
A: Cauchy Schwarz Inequality
$$\frac{\sin^3 x}{\cos x}+\frac{\cos^3 x}{\sin x}=\frac{\sin^4 x}{\cos x\sin x}+\frac{\cos^4 x}{\sin x\cos x}\geq \frac{(\sin^2 x+\cos^2 x)^2}{2\sin x\cos x}$$
$$\frac{\sin^3 x}{\cos x}+\frac{\cos^3 x}{\sin x}\geq \frac{1}{\sin 2x}\geq 1$$
equality hold when  $$\frac{\sin^2 x}{\sin x\cos x}=\frac{\cos^2 x}{\sin x\cos x}$$
that is $\displaystyle x=\frac{\pi}{4}.$
