Answer
$(-2, -1) \cup \left(-\frac{2}{3}, 0\right) $
Work Step by Step
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a Rational function.
$\displaystyle \frac{1}{x} +\frac{1}{x+1} \lt \frac{2}{x+2}$,
$\displaystyle \frac{1}{x} +\frac{1}{x+1}- \frac{2}{x+2}\lt 0$,
$\displaystyle \frac{(x+1)(x+2)+x(x+2)-2(x(x+1))}{x(x+1)(x+2)}\lt 0$,
$\displaystyle \frac{x^2+2x+x+2+x^2+2x-2x^2-2x}{x(x+1)(x+2)}\lt 0$,
$\displaystyle \frac{3x+2}{x(x+1)(x+2)}\lt 0$,
$f(x)=\displaystyle \frac{3x+2}{x(x+1)(x+2)}\lt 0$,
2.The cut points are:
$\displaystyle \frac{3x+2}{x(x+1)(x+2)} =0$
$x=-\frac{2}{3}$ or $x=0$ or $x=-1$ or $x=-2$
3. Make a table or diagram: use the test values to make a table or diagram of the sign of each factor in each interval.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \lt 0 ? \\
& & \displaystyle \displaystyle \frac{3a+2}{a(a+1)(a+2)} & \\
(-\infty,-2) & -5 & \frac{(-)}{(-)(-)(-)}=(+) & F\\
(-2, -1) & -\frac{3}{2} & \frac{(-)}{(-)(-)(+)}=(-) & T\\
(-1, -\frac{2}{3}) & -\frac{3}{4} & \frac{(-)}{(-)(+)(+)}=(+) & F\\
(-\frac{2}{3}, 0) & -\frac{1}{2} & \frac{(+)}{(-)(+)(+)}=(-) & T\\
(0,\infty) & 5 & \frac{(+)}{(+)(+)(+)}=(+) & F\\
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
Solution set: $(-2, -1) \cup \left(-\frac{2}{3}, 0\right) $
