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I'm looking for hints on how to efficiently solve this inequality: $$\left( \frac {|x|-|1-x|}{|x|} \right)^{2x-1} \gt \left(\frac {|x|-|1-x|}{|x|} \right)^{8-x} $$

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I think you may want to separate for two cases: $x>0$ and $x<0$ –  Elimination Jul 30 at 10:44
@elimination Actually probably also in $1-x>0$ and $1-x<0$ –  dreamer Jul 30 at 10:56
Right, thank you @dreamer –  Elimination Jul 30 at 10:57
We're in the process of removing the homework tag, so I've removed it from your question. –  anorton Jul 30 at 11:16

2 Answers 2

Set $y = \dfrac{|x|-|1-x|}{|x|}= 1 - \lvert \frac1x-1 \rvert$.

Note $y > 1 $ is not possible, and $y \in [0, 1]$ when $x \ge \frac12$. Otherwise $y$ is negative, (so what?).

So the only case to consider is $y \in [0, 1]$, which means $2x-1< 8-x \implies ...$

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Hahah, cough don't look at the edit history cough ;) –  dreamer Jul 30 at 11:15
@dreamer started writing a hint and ended up with short answer, so had to edit :( –  Macavity Jul 30 at 11:18
Haha, I know, I was just kidding :) –  dreamer Jul 30 at 11:22
exponential function wlays has a base whihc is greater then zero and not equal to one, and hence your said first lmitation (right?) –  Bak1139 Jul 30 at 11:47
@Bak1139 $f(x)=1^x = 1$ is trivially a constant, though technically a valid function. In the above problem $y=1$ is not a solution purely because the inequality is strict. –  Macavity Jul 30 at 12:00

One can distinguish multiple cases

  • $x>0, \ 1-x>0$
  • $x>0, \ 1-x<0$
  • $x<0, \ 1-x>0$
  • $x<0, \ 1-x<0$

As an example,when $x<0$ and $1-x<0$ we have that the inequality reads $\left(\frac{1-x-x}{-x}\right)^{2x-1}>\left(\frac{1-x-x}{-x}\right)^{8-x}$ or equivalently $\left(\frac{2x-1}{x}\right)^{2x-1}>\left(\frac{2x-1}{x}\right)^{8-x}$

It then follows that this is true for $2x-1>8-x$ or equivalently $3x>9 \rightarrow x>3$, but this is not possible since we found this solution under the assumption that $x<0$.

You could do the same for each case. This is the most intuitive solution, but Macavity's way is nicer.

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