# Increase and minimum/maximum of simple function

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?

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

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

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

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