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?

$$\frac1{1+x}-\frac x{x-1}=-\frac{x^2+1}{(x-1)(x+1)}>0$$
Hence, $$(x-1)(x+1)<0.$$
