Minima of $f(x)$ 
The miniumum value of $\left(1+ \dfrac{1}{\sin^n \alpha}\right)\left(1+ \dfrac{1}{\cos^n\alpha}\right)$ is? 

Attempt: 
I expanded the brackets and then differentiated and set the derivative equal to zero but it gets really complicated with that because of too many trig functions. 
What is the efficient way to solve it? 
The answer given is: 

 $(1+2^{\frac n2})^2$

 A: Cauchy Schwarz Inequality
$\displaystyle (a^2+b^2)(c^2+d^2)\geq (ac+bd)^2$
and equality hold when $\displaystyle \frac{a}{c} = \frac{b}{d}.$
Here I am assuming $\displaystyle \alpha \in \bigg(0,\frac{\pi}{2}\bigg).$
$$\Bigg[1+\frac{1}{\left(\sin^{\frac{n}{2}}\alpha\right)^2}\Bigg]\cdot \Bigg[1+\frac{1}{\left(\cos^{\frac{n}{2}}\alpha\right)^2}\Bigg]\geq \bigg[1+\frac{1}{\sin^{\frac{n}{2}}\alpha\cdot \cos^{\frac{n}{2}}\alpha}\bigg]^2$$
$$ = \bigg[1+\frac{2^{\frac{n}{2}}}{\left(\sin 2\alpha\right)^{\frac{n}{2}}}\bigg]^2 \geq \bigg(1+2^{\frac{n}{2}}\bigg)^2.$$
A: hint
If $n=2p$ then put
$$x=\sin^2 \alpha$$ and find the minimum of
$$\left(1+\frac{1}{x^p}\right)\left(1+\frac{1}{(1-x)^p}\right)$$
A: Let $$f(x)=\left(1+\frac{1}{\sin^nx}\right)\left(1+\frac{1}{\cos^nx}\right)\text{ for } 0<x<\frac{\pi}{2}$$
Then $$\begin{align}
f(x)&=\left(1+\frac{1}{\sin^nx}\right)'\left(1+\frac{1}{\cos^nx}\right)+
\left(1+\frac{1}{\sin^nx}\right)\left(1+\frac{1}{\cos^nx}\right)'\\
&=\frac{-n\cos x}{\sin^{n+1}x}\left(1+\frac{1}{\cos^nx}\right)+
\frac{n\sin x}{\cos^{n+1}x}\left(1+\frac{1}{\sin^nx}\right)\\
&=\frac{n}{\sin^nx\cos^nx}\left(-\frac{\cos^{n+1}x+\cos x}{\sin x}+\frac{\sin^{n+1}x+\sin x}{\cos x}\right)
\end{align}
$$
Since $\cos x > \sin x > 0$ on $(0, \pi/4),$ we see that $f'(x)<0$  on $(0,\pi/4).$  Similarly, $f'(x)>0$ on $(\pi/4,\pi/2),$ so that $f$ attains it minimum on $(0,\pi/2)$ at $x=\pi/4$.
