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$\frac{x-1}{x+1} < x$


I did the following.

$\frac{x-1}{x+1} - x< 0 /-x$

$\frac{x-1 - x(x+1)}{x+1} < 0$

$\frac{-x^2-1}{x+1} < 0$

What to do next?

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I did the following. – George May 25 '14 at 20:08
Never mind...I got it. :D – George May 25 '14 at 20:13

Note that the inequality can be written as

$$-(x^2 + 1) \cdot \frac{1}{x + 1} < 0$$

Now $x^2 + 1$ is always positive, so this is equivalent to studying

$$\frac{1}{x + 1} > 0$$

Can you solve this?

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Yes, thanks! :) – George May 25 '14 at 20:15
You're very welcome. – user61527 May 25 '14 at 20:15

Start with: $x\le\frac{1}{2}(x^2+1)\Rightarrow x-1\le\frac{1}{2}(x^2-1)$

Thus $\frac{x-1}{x+1}\le\frac{1}{2}(x-1)\le x-1\lt x$

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@achillehui yes you are right.. – ellya May 25 '14 at 20:21

For $x>-1$ the inequality is equivalent to:



Hence we have a solution for all x in this interval.

For $x<-1$ the inequality is equivalent to:



Thus we have no solutions in this interval.

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$\frac{x-1}{x+1}=\frac{(x+1)-2}{x+1}=1+\frac{-2}{x+1}<x$. Multiply both sides by $x+1$ and simplify what you get. I think that's the easiest way to simplify this inequality.

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Nope. That's not a valid step because $x+1$ varies in sign. Doing something like that requires you to produce a separate inequality for each case of the sign of the multiplier (positive, negative, and zero). – MPW May 25 '14 at 20:23
Honestly thats what i would have done. I now realize that there are better ways to answer it. – Asimov May 25 '14 at 20:51

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