A calculus problem by Ossama Ismail

We note that $\dfrac {\sin^3 x}{\cos x} > 0$ and $\dfrac {\cos^3 x}{\sin x} > 0$ for $0 < x < \frac \pi 2$ . We can thus apply AM-GM inequality as follows:

$\begin{aligned} \frac {\sin^3 x}{\cos x} + \frac {\cos^3 x}{\sin x} & \ge 2 \sqrt {\frac {\sin^3 x}{\cos x} \cdot \frac {\cos^3 x}{\sin x}} = 2 \sin x \cos x = \sin (2x) \end{aligned}$

Equality occurs when $\dfrac {\sin^3 x}{\cos x} = \dfrac {\cos^3 x}{\sin x} \implies \sin^4 x = \cos^4 x \implies x = \frac \pi 4$

$\begin{aligned} \frac {\sin^3 x}{\cos x} + \frac {\cos^3 x}{\sin x} & \ge \sin \frac \pi 2 = \boxed{1} \end{aligned}$

We can solve this problem by differentiation, i.e. df/dx = 0. for f(x) = (sin ^4 x+cos ^4 x)/(sin⁡ x. cos⁡x ) f(x) = (sin^4 x+cos^4 x)/(sin⁡x. cos⁡x ) = (1-2sin^2 xcos^2 x)/(1/2 sin⁡2x ) = (1-1/2 sin^2 2x)/(1/2 sin⁡2x ) = (2/sin⁡2x) - sin 2x df/dx = - cos 2x (4/(〖sin〗^2 2x) + 2), df/df=0, cos 2x= 0, x=π/4, Substitute x=π/4, to the expression above then we find the minimum value is 1