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.
 A: If I may suggest, consider $$f(x)=\frac{1}{x}-\frac{1}{x-1}-\frac{1}{x-2}$$ that you want to be positive. Reduce to same denominator and simplify; you then arrive to $$f(x)=\frac{2-x^2}{x(x-1)(x-2)}$$ Multiply the top by the denominator so $$f(x)=\frac{(2-x^2)x(x-1)(x-2)}{\Big(x(x-1)(x-2)\Big)^2}$$ Now, the denominator is positive and so the sign of $f(x)$ is given by the sign of $$(\sqrt 2-x)(\sqrt 2+x)x(x-1)(x-2)$$
I am sure that you can take from here.
A: Combining all the fractions on one side of the inequality and use "test points" to determine the intervals of validity.
$$\frac{1}{x}-\frac{1}{x-1}>\frac{1}{x-2}$$
$$\frac{1}{x}-\frac{1}{x-1}-\frac{1}{x-2}>0$$
$$\frac{2-x^2}{x(x-1)(x-2)}>0$$
The rational function can only change sign near a vertical asymptote or a zero.
The zeros are $\pm\sqrt{2}$, and the vertical asymptotes are at 0, 1 and 2. Since we are only looking for positive values, possible intervals of validity are $(0,1),(1,\sqrt{2}),(\sqrt{2},2),(2,\infty).$ Pick a member of each interval and test whether it satisfies the inequality. Example: you said that .25 works in the original inequality. Since .25 belongs to the interval (0,1) the entire interval (0,1) is valid. Test points in the other intervals, and unite the collection of valid intervals to create the solution set.
A: 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 $(0,1)\cup(\sqrt{2},2)$. 
