Positive values of $x$ that satisfy the inequality $\frac{1}{x}-\frac{1}{x-1}>\frac{1}{x-2}$ Determine the set of positive values of $x$ that satisfy the inequality $$\frac{1}{x}-\frac{1}{x-1}>\frac{1}{x-2}.$$
My attempt:
\begin{align}
\frac{-1}{x(x-1)} & >\frac{1}{(x-2)} \\[0.1in]
\frac{1}{x(1-x)} & >\frac{1}{(x-2)} \\[0.1in]
 x(1-x) & <(x-2)
\end{align}
If I put $0.25$ in the original inequality, it works, but not in the last one. What mistake did I make? Please advise.

First simplify the inequality:
\begin{align}
\frac{1}{x} - \frac{1}{x-1} > \frac{1}{x-2} & \implies \frac{1}{x} - \frac{1}{x-1} -\frac{1}{x-2} > 0 \\[0.1in]
& \implies \frac{(\sqrt{2}-x)(\sqrt{2}+x)}{x(x-1)(x-2)} > 0 \\[0.1in]
\end{align}
So the zeros of the numerator and denominator, listed in order, are $-\sqrt{2},0,1,\sqrt{2},2$. We only need to see if $$f(x) = \frac{(\sqrt{2}-x)(\sqrt{2}+x)}{x(x-1)(x-2)}$$ is positive on which of the intervals $(-\infty,-\sqrt{2}),(-\sqrt{2},0),(0,1),(1,\sqrt{2}),(\sqrt{2},2),(2,\infty)$. Within each of these intervals, the function $f(x)$ is defined and can never be zero. Therefore we only need to test one point from each interval to determine if $f(x)$ is positive on it. If I am correct, then it should be $(-\infty,-\sqrt{2}) \cup (0,1)\cup(\sqrt{2},2)$. 
Now if you only want positive values of $x$, then it should be $(0,1)\cup(\sqrt{2},2)$. 
