<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] 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.
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