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

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 $\displaystyle \sin^n \alpha = \cos^n \alpha\Rightarrow \alpha = \frac{\pi}{4}$
