<s>[INST] 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
 [/INST] 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 $\displaystyle \sin^n \alpha = \cos^n \alpha\Rightarrow \alpha = \frac{\pi}{4}$
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