<s>[INST] 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?
 [/INST] 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?
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