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?

Writing your inequality in the form
$$\frac{1}{x+1}-\frac{x}{x-1}>0$$ this is 
$$-\frac{x^2+1}{x^2-1}>0$$ or $$\frac{x^2+1}{x^2-1}<0$$
Can you finish?
The only solution we get for $$x^2-1<0$$ since $$x^2+1>0$$ holds for all real numbers.
