# 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.

• You're probably right. But there is a way without using calculus. You should only use Cauchy's inequality: $$\frac{a_1+a_2}{2}\geq \sqrt{a_1a_2}.$$ Sep 23, 2018 at 13:25

You're probably right. But there is a way without using calculus. You should only use Cauchy's inequality:

$$\frac{a_1+a_2}{2}\geq \sqrt{a_1a_2}.$$

$$\left(\frac{1+\sin^2x}{\sin^2x}\right)^n+\left(\frac{1+\cos^2x}{\cos^2x}\right)^n=\left(1+\frac{1}{\sin^2x}\right)^n+\left(1+\frac{1}{\cos^2x}\right)^n \geq \\ \geq 2\left(\sqrt{\left(1+\frac{1}{\sin^2x}\right)\left(1+\frac{1}{\cos^2x}\right)}\right)^n=2\left(\sqrt{1+\frac{1}{\sin^2x}+\frac{1}{\cos^2x}+\frac{1}{\sin^2x\cos^2x}}\right)^n= \\ = 2\left(\sqrt{1+\frac{2}{\sin^2x\cos^2x}}\right)^n=2\left(\sqrt{1+\frac{8}{\sin^22x}}\right)^n\geq 2\left(\sqrt{1+8}\right)^n=2\cdot3^n$$

We must now prove there is $$x_1$$ such that $$f(x_1)=2\cdot3^n$$.

To that end, notice that $$f(x)=2\cdot3^n$$ is equivalent to following system

$$\begin{cases} \sin^2x=\cos^2x, \\ \sin^22x=1, \end{cases}$$

which has $$x_1=\pi/4$$ as solution. Now note that

$$f\left(\frac{\pi}{4}\right)=\left(\frac{1+1/2}{1/2}\right)^n+\left(\frac{1+1/2}{1/2}\right)^n=2\cdot3^n.$$

Thereby, the minimum value of $$f(x)$$ is

$$f_{min}=2\cdot3^n.$$

• ooh! Thank you for your solution. Do you think it's possible to do this by derivative?
– user596235
Sep 23, 2018 at 14:34
• @Ryany I think so, but I'm in between two positions at the moment. I will try it as soon as possible but it might take some time. Sep 23, 2018 at 14:42

Using A.M $$\geq G.M.$$ on the first and second inequalities yields

\begin{aligned} f(x) &=\left(\csc ^{2} x+1\right)^{n}+\left(\sec ^{2} x+1\right)^{n} \\ & \geqslant 2 \sqrt{\left[\left(\csc ^{2} x+1\right)\left(\sec ^{2} x+1\right)\right]^{n}} \\ &=2\left[\left(2+\cot ^{2} x\right)\left(2+\tan ^{2} x\right)\right]^{\frac{n}{2}} \\ &=2\left(4+2 \cot ^{2} x+2 \tan ^{2} x+1\right)^{\frac{n}{2}} \\ & \geqslant 2\left(4+2 \cdot 2 \sqrt{\cot ^{2} x \tan ^{2} x}+1\right)^{\frac{n}{2}} \\ &=2\cdot 9^{\frac{n}{2}} \\ &=2\cdot 3^{n} \end{aligned} $$\therefore f(x)$$ attains its minimum value $$2 \cdot 3^{n}$$ when $$\cot ^{2} x=\tan ^{2} x \textrm{ and } \csc^2 x=sec^2 x\Leftrightarrow x=\pm \frac{\pi}{4}.$$

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 $$x = \dfrac{\pi}{4}$$.