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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.

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    @Karatug: The inequality is valid for $x=\sqrt{3}$ and $\sqrt{3}\in(0,1)\cup(\sqrt{2},2)$ (the "given answer"). So you're saying *OP's* answer is wrong, correct?2011-11-06

3 Answers 3

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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 $02$. But we are not OK if $1

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    All 3 answers are really helpful, but I think this one was the most helpful. Thanks a lot!2011-11-07
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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 $x,x-1,x-2$.

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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

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 $f(x)0$.