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?
 A: Hint: You have to be careful multiplying both sides of an inequality by a quantity that depends on a variable. The expression could be positive, or the quantity could be negative (how does this affect the direction of the inequality?).
A: 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.
A: $$\frac{1}{x+1}-\frac{x}{x-1}=\frac{-x^2-1}{(x-1)(x+1)}.$$
Thus, $(x-1)(x+1)<0,$ which gives $-1<x<1.$
A: $$\frac1{1+x}-\frac x{x-1}=-\frac{x^2+1}{(x-1)(x+1)}>0$$
Hence, $$(x-1)(x+1)<0.$$
A: 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 $-1<x<1$. 
