Solve the inequality $\frac 1{x+1} - \frac 1{x} \le \frac 1{x-1} - \frac 1{x-2}$ $$\frac 1{x+1} - \frac 1{x} \le \frac 1{x-1} - \frac 1{x-2}$$
The answer for this inequality is given as $x ∈ (- \infty , -1) \cup(0, 1) \cup (2, \infty)$ but when I solve it, I am only able to get $x ∈ ( - \infty, -1)$.
How should I solve it to get the complete answer?
 A: Go step by step:
$$\frac 1{x+1} - \frac 1{x} \le \frac 1{x-1} - \frac 1{x-2} \\
\frac{-1}{x(x+1)} \le \frac{-1}{(x-2)(x-1)} \\
\frac{1}{(x-2)(x-1)} - \frac{1}{x(x+1)} \le 0 \\
\frac{x(x+1) - (x-2)(x-1)}{(x-2)(x-1)x(x+1)} \le 0 \\
\frac{x^2+x - (x^2-3x+2)}{(x-2)(x-1)x(x+1)} \le 0 \\
\frac{4x-2}{(x-2)(x-1)x(x+1)} \le 0$$
Can you finish this now?
A: Solve the Inequality :$$\frac 1{x+1} - \frac 1{x} \le \frac 1{x-1} - \frac 1{x-2} \\$$
$$\frac{-1}{x(x+1)} \le \frac{-1}{(x-2)(x-1)} \\$$
Consider two scenarios :
scenario 1 : Less than
$$\frac{-1}{x(x+1)}~< \frac{-1}{(x-2)(x-1) }$$.
$$\frac{-1~(x^2-x-2x+2)}{x(x+1)}+1~<0$$
$$\frac{(4~x~-2)}{x~(~x+1)}~<0$$
$$x \in (-\infty , \frac{1}{2}) $$
Scenario 2 : equal to
$$\frac{-1}{x(x+1)}~=\frac{-1}{(x-2)(x-1) }$$.
$$\frac{-1~(x^2-x-2x+2)}{x(x+1)}+1~=0$$
$$\frac{(4~x~-2)}{x~(~x+1)}~=0$$
$$ x = \frac{1}{2}$$
Final step : $$x \in (-\infty , \frac{1}{2}] $$
A: Another way to tackle this problem:
\begin{align}\frac 1{x+1} - \frac 1{x}=-\frac 1{x(x+1)} &\le \frac 1{x-1} - \frac 1{x-2}= -\frac 1{(x-2)(x-1)}\\
\iff \frac{1}{(x-2)(x-1)}&\le  \frac{1}{x(x+1)} 
\end{align}
Now observe that:

*

*outside of $\,(-1,0)\cup(1,2)$, both denominators are positive, so
$$\frac{1}{(x-2)(x-1)}\le  \frac{1}{x(x+1)}\iff x(x+1)\le(x-2)(x-1)\iff 4x\le 2.$$

*In $(-1,0)$ and in $(1,2)$, compare the signs of the denominators.

