Answer
$(-\infty,-3-\sqrt 2]\cup(-3,-2)\cup[-3+\sqrt 2,-1)$
Work Step by Step
1. Move all items to one side $f(x)=\frac{1}{x+1}+\frac{1}{x+2}-\frac{1}{x+3}
=\frac{x^2+5x+6+x^2+3x+3-x^2-2x-2}{(x+1)(x+2)(x+3)}
=\frac{x^2+6x+7}{(x+1)(x+2)(x+3)}$
$f(x)==\frac{(x+3+\sqrt 2)(x+3-\sqrt 2)}{(x+1)(x+2)(x+3)}\leq0$
2. Cut points $-3-\sqrt 2 (-4.4), -3,-2,-3+\sqrt 2 (-1.6),-1$, make a table as shown.
3. Write the solutions $(-\infty,-3-\sqrt 2]\cup(-3,-2)\cup[-3+\sqrt 2,-1)$