Minimum of $\left(\frac{1+\sin^2x}{\sin^2x}\right)^n+\left(\frac{1+\cos^2x}{\cos^2x}\right)^n$ 
I would like to find the minimum of
$$f(x)=\left(\frac{1+\sin^2x}{\sin^2x}\right)^n+\left(\frac{1+\cos^2x}{\cos^2x}\right)^n,$$
where $n$ is a natural number.

I know there is possible by derivate, but
$$f'(x)=n \left(\left(\cos ^2(x)+1\right) \sec ^2(x)\right)^{n-1} \left(2 \left(\cos ^2(x)+1\right)\tan (x) \sec ^2(x)-2 \tan (x)\right)+n \left(\left(\sin ^2(x)+1\right) \csc^2(x)\right)^{n-1} \left(2 \cot (x)-2 \left(\sin ^2(x)+1\right) \cot (x) \csc^2(x)\right).$$
I think this is not the best way.

Let $A = \dfrac{1+\sin^2x}{\sin^2x}, B = \dfrac{1+\cos^2x}{\cos^2x}$. Then $A+B = 2+\dfrac{1^2}{\sin^2x}+\dfrac{1^2}{\cos^2x}\ge 2+\dfrac{(1+1)^2}{\sin^2x+\cos^2x}=6$. Thus: $A^n+B^n\ge \dfrac{(A+B)^n}{2^{n-1}}= \dfrac{6^n}{2^{n-1}}=2\cdot 3^n$. Equality occurs when $\sin x = \cos x$ or $x = \dfrac{\pi}{4}$.
