<s>[INST] 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.
 [/INST] 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.
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