Minimum value of $(\sin x + \cos x)^3 + \frac{1}{\sin^2x\cos^2x}$

Question

Let $\sin x + \cos x = t\qquad (\vert t \vert \leq \sqrt2)$

and solve it as $t^3 + \dfrac{4}{(t^2 - 1)^2}$

Is there any easier solution to solve this $(\sin x + \cos x)^3 + \dfrac{1}{\sin^2x\cos^2x}$?

Is there any available inequality can solve it?

Please help!

Answer 1

There may not be a specific inequality, but sure there is a faster way.

$$(\sin x + \cos x)^3 + \dfrac{1}{\sin^2x\cos^2x}=2\sqrt{2}\left(\sin\left(x+\frac{\pi}{4}\right)\right)^{3}+4\csc^{2}\left(2x\right)$$

Observe that only the expression having an odd power can contribute to reducing the value of the expression. That is, when $\sin(x+\pi/4)$ is the lowest.

A possible solution is $x=-\dfrac{3\pi}{4}$ which gives the global minimum as $4-2\sqrt{2}=1.1715728\dots$

Here is a graphical solution in aid of the above.