<s>[INST] How to solve this weird inequality? $\frac{x-1}{x+1} < x$
Thanks!
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?
 [/INST] 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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