<s>[INST] Find real $x$ that satisfies $\frac{1}{1+x}>\frac{x}{x-1}$ $\dfrac{1}{1+x}>\dfrac{x}{x-1}$
Multiply both sides with $(x+1)(x-1)$.
$x-1>x(x+1)$
Subtract $(x-1)$ from both sides.
$0>x^2+1$
This seems to have no real answers, although I have been told there should be. What am I doing wrong?
 [/INST] $$\frac{1}{x+1}-\frac{x}{x-1}=\frac{-x^2-1}{(x-1)(x+1)}.$$
Thus, $(x-1)(x+1)<0,$ which gives $-1<x<1.$
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