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?

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 : $$x \in (-\infty , \frac{1}{2}] $$
