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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?


  • 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
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$x^2(1-x) > 0$ for $x ∈ (0..1)$, so instead show $x^2+1 > 8x^2(1-x) = 8x^2 - 8x^3$. – k.stm Mar 10 '14 at 13:07
Maybe rewriting LHS as $\frac{1}{1-\frac{x^3 + 1}{x^2 + 1}}$ might help. – Yiyuan Lee Mar 10 '14 at 13:09
Alternatively: factor the denominator as $x \left(x (1-x) \right)$. Then show that $x(1-x)$ is no greater than 1/4 on the interval. – John Hughes Mar 10 '14 at 13:09
@k.stm I have $(2x+1)(4x^2-2x+1)>7x^2$ am I on good way? because I don't know what next – Gregor Mar 10 '14 at 13:16
@Gregor I expanded this idea into a full answer. – k.stm Mar 10 '14 at 13:54
up vote 1 down vote accepted

Hint: Show $p(x) = 8x^3 - 7x^2 + 1 > 0$ for all $x ∈ (0..1)$ by showing that $p$ has a positive minimum on $[0..1]$. Look at $p(0)$, at $p(1)$ and at $p'$ on $(0..1)$.

I have carried this out below.

First observe that the only zero of $p'$ in $(0..1)$ is at $x = 7/12$, since $p'(x) = x(24x-14)$ for $x ∈ (0..1)$.

Now because $p(7/12) > 0$, the only possible local extremum (be it maximum or minimum) in $(0..1)$ is positive. Also $p(0) > 0$ and $p(1) > 0$.

On the compact intervall $[0..1]$, the continuous function $p$ assumes both maximum and minimum, but as just shown, neither at $0$ nor at $1$ nor somewhere in between on $(0..1)$ a nonpositive minimum can be achieved. So the minimum must be positive and therefore $p > 0$ on all $[0..1]$.

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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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