Solving $\frac1{x-2} + \frac1{x-1} > \frac1x$

Question

$$\frac1{x-2} + \frac1{x-1} > \frac1x$$

My attempt:

I solved it and got $\frac{-x^2+5x-5}{(x-2)(x-1)(x) }> 0$ But after this I am unable to proceed as the numerator has no real roots. So please guide me how shall I proceed.

Answer 1

$x \notin \{0,1,2\}$,

$$\frac1{x-2}+\frac1{x-1} > \frac1x$$

• If $x>2$, then each terms are positive, then clearly it is true.

Now, we focus on $x < 2$, that is $x-2<0$.

$$\frac1{x-1}-\frac1x > \frac1{2-x}$$

$$\frac{2-x}{x(x-1)}>1$$

$$\frac{2-x-x(x-1)}{x(x-1)}>0$$

$$\frac{2-x^2}{x(x-1)}>0$$

• If $2-x^2 > 0$, then $x(x-1)>0$. that is $ -\sqrt2 < x < \sqrt2$ and ($x<0$ or $x>1$). That is we obtain $(-\sqrt2, 0) \cup (1, \sqrt2)$.

• If $2-x^2 < 0$, then $x(x-1) <0$, that is $|x| > \sqrt2$ and $0\le x \le 1$, of which no such value exists.

Hence in summary, $(-\sqrt2, 0) \cup (1, \sqrt2)\cup (2, \infty)$.

Answer 2

Actually,$$\frac1{x-2}+\frac1{x-1}-\frac1x=\frac{x^2-2}{x(x-1)(x-2)}.$$Besides:

• $x^2-2>0$ if and only if $x\in\left(-\infty,-\sqrt2\right)\cup\left(\sqrt2,\infty\right)$;
• $x^2-2<0$ if and only if $x\in\left(-\sqrt2,\sqrt2\right)$;
• $x(x-1)(x-2)>0$ if and only if $x\in(0,1)\cup(2,\infty)$;
• $x(x-1)(x-2)<0$ if and only if $x\in(-\infty,0)\cup(1,2)$.

Can you take it from here?

Answer 3

The inequality becomes $$\frac1{x-2}+\frac1{x-1}-\frac1x=\frac{x^2-2}{x(x-1)(x-2)}>0~~~~(1)$$ $$ F=Sign \frac{x^2-2}{x(x-1)(x-2)} = Sign [(x+\sqrt{2})x(x-1)(x-\sqrt{2})(x-2)]~~~~(2)$$ For real values of $x$ (1) and (2) are equivalent to $F>0$ by wavy curve method we get that all real values of $x$ are given by: $$ (-\sqrt{2};0) \cup (1;\sqrt{2}) \cup (2; \infty)$$