Minimum of $ f(\alpha) = \left(1+\frac{1}{\sin^{n}\alpha}\right)\cdot \left(1+\frac{1}{\cos^{n}\alpha}\right)$ 
Minimum value of $\displaystyle f(\alpha) = \left(1+\frac{1}{\sin^{n}\alpha}\right)\cdot \left(1+\frac{1}{\cos^{n}\alpha}\right)\;,$ Where $n\in \mathbb{N}$ and $\displaystyle \alpha \in \left(0,\frac{\pi}{2}\right)$

I have solved It using Derivative Test, But that nethod is very Lengthy,
Can we solve it Using Inequality
If yes ,Then plz explain me how can we solve it, Thanks
 A: The following is not a complete answer, but it proves that $f'(\pi/4)=0$ without actally differetiating $f$. I think this should help.
Define for $x\in(-\pi/4,\pi/4)$
$$g(x)=f(x-\pi/4)=\left(1+\frac1{(\frac{\sqrt2}2\sin x-\frac{\sqrt2}2\cos x)^n}\right)\left(1+\frac1{(\frac{\sqrt2}2\sin x+\frac{\sqrt2}2\cos x)^n}\right)$$
Note that
$$g(-x)=(-1)^n\left(1+\frac1{(\frac{\sqrt2}2\sin x+\frac{\sqrt2}2\cos x)^n}\right)(-1)^n\left(1+\frac1{(\frac{\sqrt2}2\sin x-\frac{\sqrt2}2\cos x)^n}\right)=g(x)$$
Therefore, $g$ is an even function. Since $g$ is differentiable at $0$, we have $g'(0)=0$.
A: Thanks friends  got it
Let $\displaystyle f(x) = \ln\left(1+\frac{1}{t}\right)\;, t>0$ Using Jesan Inequality function $f(t)$ is convex function.
So $$\ln\left(1+\frac{1}{x}\right)+\ln\left(1+\frac{1}{y}\right)\geq 2\ln\left(1+\frac{2}{x+y}\right)$$
Where $x,y\in (0,1)$
So $$\ln\left(1+\frac{1}{x}\right)+\ln\left(1+\frac{1}{y}\right)\geq \ln\left(1+\frac{2}{x+y}\right)^2\geq \ln \left(1+\frac{1}{\sqrt{xy}}\right)^2$$
So $$\left(1+\frac{1}{x}\right)\cdot \left(1+\frac{1}{y}\right)\geq \left(1+\frac{1}{\sqrt{xy}}\right)^2$$ and equality hold when $x=y$
So $$\left(1+\frac{1}{\sin^ n\alpha}\right)\cdot \left(1+\frac{1}{\cos^n \alpha}\right)\geq \left(1+\frac{2^{\frac{n}{2}}}{\sqrt{\sin 2 \alpha}}\right)^2\geq \left(1+2^{\frac{n}{2}}\right)^2$$
and equality hold when $\displaystyle \sin^n \alpha = \cos^n \alpha\Rightarrow \alpha = \frac{\pi}{4}$
