<s>[INST] Prove that $\frac{x^2+1}{x^2(1-x)}>8$ for $x\in(0,1)$ 
Show that $\displaystyle \frac{x^2+1}{x^2(1-x)}>8$ for $x\in(0,1)$

I thought about derivative but I think it's too complicated, do you have any ideas?
Progress

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*instead show $x^2+1 > 8x^2(1-x) = 8x^2 - 8x^3$. -- k.stm

*I have $(2x+1)(4x^2-2x+1)>7x^2$ am I on good way? because I don't know what next

 [/INST] We have, $\displaystyle \frac{x^2+1}{x^2(1-x)} = \displaystyle \frac{x+\frac{1}{x}}{x(1-x)}= \displaystyle \frac{2 + (\sqrt x - \frac{1}{\sqrt x})^2}{\frac{1}{4} - (x-\frac{1}{2})^2} > 8$.
(inequality is strict since $x\in(0,1)$).
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