1/x+1/x+2≤1/x+1
Geherally the best way to solve a complicated inequality, like $\displaystyle \frac{1}{x}+ \frac{1}{x+2}\le \frac{1}{x+1}$ or, equivalently $\displaystyle \frac{1}{x}+ \frac{1}{x-2}-\frac{1}{x+1}\le 0$, is to solve the corresponding equation: $\displaystyle \frac{1}{x}+ \frac{1}{x+2}= \frac{1}{x+1}$.
Multiply the entire equation by x(x+1)(x+2): (x+1)(x+2)+ x(x+1)= x(x+2). Multiplying that out, $\displaystyle x^2+ 3x+ 2+ x^2+ x= x^2+ 2x$ which reduces to $\displaystyle x^2+ x+ 2= 0$. The discriminant for that is [tex]2^2- 4(1)(2)= -4 so there are NO real number roots.
A function can change from ">" to "<" or vice-versa only at points where it is "=" or where it is discontinuous. Since there are no points where it is "=", it can only change at points where it is not continuous: where one of the denominators is 0, that is at x= -1, x= -2, and x= 0.
You can check a single value in each interval having those endpoints to determine whether every point in that interval satisfies the inequality or not.
For example, -3 is in the interval of all numbers less than -2 and [tex]\frac{1}{-3}+ \frac{1}{-3+ 2}= -\frac{1}{3}-1= -\frac{4}{3} math which < is than [tex]\frac{1}{-3+1}= -\frac{1}{2}[tex] so the inequality is satisified for all x< -2. Try, say, x= -3/2, which is between -2 and -1 to check if that interval works, x= -1/2 for the interval (-1, 0) and x= 1 for the interval of numbers larger than 1.