Manipulating this Inequality What's wrong with this method of trying to find the set of positive values of $x$ that satisfy the following inequality:

$$\dfrac{1}{x}-\dfrac{1}{x-1} > \dfrac{1}{x-2}$$

Find a common denominator on the left hand side:
$\dfrac{-1}{x(x-1)}>\dfrac{1}{x-2}\Leftrightarrow$
$2-x > x(x-1)\Leftrightarrow$
$2 > x^2$
Which is satisfied whenever $0 < x < \sqrt{2}$ (we only want positive $x$).
From our original inequality, we throw out the points $x=0, x=1, x=2$. But for $\sqrt{2} < x < 2$ we don't satisfy $x^2 < 2$, so I thought it should be $(0,1)\cup(1,\sqrt{2})$. This is wrong though... answer is $(0,1)\cup(\sqrt{2},2)$. Hope someone can help explain to me what steps were wrong. Maybe tell me explicitly which of the if and only if implications don't hold. Thanks a lot.
 A: You can't multiply $$\frac{-1}{x(x-1)}>\frac{1}{x-2}$$
through by $x-2$ and $x(x-1)$ unless both of those are positive. If we know $x>0$ then split the situation up into three cases: $x\in(0,1)$, $x\in(1,2)$ and $x\in(2,\infty)$ (note $x\in\{1,2\}$ is impossible).
Remember that if we have e.g. $a>b$ then multiplying by $-1$ gives $-a<-b$. Similarly for other negative numbers as well; use this while keeping track of the signs of $x,x-1,x-2$.
A: It is not true that 
$$\dfrac{-1}{x(x-1)}>\dfrac{1}{x-2}\Leftrightarrow 2-x > x(x-1)$$
If $c>0$, then $a>b$ is equivalent to $ac>bc$.  If $c<0$, then $a>b$ is equivalent to $ac<bc$.  Here you multiplied by $x(x-1)(x-2)$ and assumed it is positive, but this depends on $x$. That expression will not be positive when $x<0$, but you said you are ignoring that case anyway.  The expression is also negative when $1<x<2$, so you'll have the inequality backwards there.
Another way to handle this (instead of considering cases when multiplying) is to check between all of the "critical points" of the inequality.  To solve an inequality of the form $f(x)>g(x)$ where $f$ and $g$ are rational functions, note that the only points where $f(x)>g(x)$ can change to $f(x)<g(x)$ (or vice versa) are at those values of $x$ where $f(x)=g(x)$, or where one of $f(x)$ or $g(x)$ is undefined.  So you solve the "related equation" $f(x)=g(x)$ to get $x=\pm\sqrt 2$, and also take note of the points where one of the expressions is undefined, namely $x=0,1,$ or $2$.  The full list of critical points is $\{-\sqrt 2, 0,1,\sqrt 2,2\}$, and you can finish by testing a single point in each interval between the critical points.  That would be $6$ intervals to test altogether, but only $4$ if you are assuming $x>0$.
A: You cleared denominators. Think of the process as multiplying both sides by $x(x-1)(x-2)$.
This preserves the inequality if $x(x-1)(x-2)$ is positive, but reverses the inequality when $x(x-1)(x-2)$ is negative.
Since $x$ is positive, we need not worry about it.  If $0<x<1$, then $(x-1)(x-2)$ is positive, so we are OK. We are also OK if $x>2$.  But we are not OK if $1<x<2$. There, the inequality is reversed.  
