Answer
$(-\infty,-3-\sqrt2] \cup(-3, -2) \cup[-3+\sqrt2, -1) $
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.
$\displaystyle \frac{1}{x+1}+\frac{1}{x+2} \leq \frac{1}{x+3}$,
$\displaystyle \frac{1}{x+1}+\frac{1}{x+2} -\frac{1}{x+3} \leq 0$ ,
$\displaystyle \frac{(x+3+\sqrt2)(x+3-\sqrt2)}{(x+1)(x+2)(x+3)}\leq 0$ ,
$f(x)=\displaystyle \frac{(x+3+\sqrt2)(x+3-\sqrt2)}{(x+1)(x+2)(x+3)}\leq 0$,
2.The cut points are:
$\displaystyle \frac{(x+3+\sqrt2)(x+3-\sqrt2)}{(x+1)(x+2)(x+3)}=0$
$x=-3+\sqrt2$ or $x=-3-\sqrt2$ or $x=-1$ or $x=-2$ or $x=-3$
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,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \leq 0 ? \\
& & \displaystyle \displaystyle \frac{(a+3+\sqrt2)(a+3-\sqrt2)}{(x+1)(a+2)(a+3)}& \\
(-\infty,-3-\sqrt2) & -5 & \frac{(-)(-)}{(-)(-)(-)}=(-) & T\\
(-3-\sqrt2, -3) & -4 & \frac{(-)(+)}{(-)(-)(-)}=(+) & F\\
(-3, -2) & -\frac{5}{2} & \frac{(-)(+)}{(-)(-)(+)}=(-) & T\\
(-2, -3+\sqrt2) & -\frac{5}{3} & \frac{(-)(+)}{(-)(+)(+)}=(+) & F\\
(-3+\sqrt2, -1) & -\frac{3}{2} & \frac{(+)(+)}{(-)(+)(+)}=(-) & T\\
(-1,\infty) & 0 & \frac{(+)(+)}{(+)(+)(+)}=(+) & F\\
\end{array}$
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: $(-\infty,-3-\sqrt2] \cup(-3, -2) \cup[-3+\sqrt2, -1) $
