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
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 $g'(0)=0$.
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 $\displaystyle \sin^n \alpha = \cos^n \alpha\Rightarrow \alpha = \frac{\pi}{4}$