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I was solving a question and had to find the minimum and maximum of the function $f(x) = \frac{1}{x\sin x}-\frac{1}{x^2}$ in the range $0 < x <\frac{\pi}{2}$. This function seems to increase and therefore gets minimum at $x\to0$ and maximum at $x=\frac{\pi}{2}$. But I couldn't prove that the function increases. I tried $f'(x) > 0$ which is equivalent with $\frac{2-x^2\cot(x)\csc(x)-x\csc(x)}{x^3} > 0$, or $x^2\cot(x)\csc(x)-x\csc(x) < 2$. But I couldn't prove this either. Can someone help me?

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HINT

We have that

$$2-x^2\cot(x)\csc(x)-x\csc(x) =2-\frac x{\sin x}\left(1+\frac{x\cos x}{\sin x}\right)$$

and

$$\frac x{\sin x}\left(1+\frac{x\cos x}{\sin x}\right)=\frac x{\sin x}+\left(\frac x{\sin x}\right)^2\cos x\le \frac x{x-\frac16 x^3}+\left(\frac x{x-\frac16 x^3}\right)^2\left(1-\frac12x^2+\frac1{24}x^4\right)=$$

$$=\frac{3(x^4-16x^2+48)}{2(x^2-6)^2}$$

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  • $\begingroup$ Thanks solved by it!! $\endgroup$ – Hypernova Dec 24 '18 at 14:55

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