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?

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.

