<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] The direction of an inequality changes if you multiply by something negative, so a common tactic is to multiply by a square. Multiplying by $(x^2-1)^2$ gives $(x-1)(x^2-1)\ge x(x+1)(x^2-1)$, i.e. $(1+x^2)(1-x^2)\ge 0$. (The $>$ has become $\ge$ in case $x^2-1=0$, although in this case the original problem precludes $x=\pm 1$ due to the divisions by $x\mp 1$.) Cancelling a positive factor, $1-x^2\ge 0$ i.e. $-1\le x\le 1$. As already noted, we have to drop the cases $x=\pm 1$, giving $-1<x<1$. 
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