$\frac{1}{x+2}>\frac{1}{x}-\frac{1}{x+1}$

$\frac{1}{x+2}-\frac{1}{x}+\frac{1}{x+1}>0\Rightarrow \frac{x\left(x+1\right)-\left(x+2\right)\left(x+1\right)+x\left(x+2\right)}{x\left(x+2\right)\left(x+1\right)}>0\Rightarrow \frac{x^2+x-x^2-x-2x-2+x^2+2x}{x\left(x+2\right)\left(x+1\right)}>0\Rightarrow$

$\Rightarrow \frac{x^2-2}{x\left(x+2\right)\left(x+1\right)}>0\Rightarrow \frac{\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)}{x\left(x+2\right)\left(x+1\right)}>0\Rightarrow x\in ]-2,-\sqrt{2}[\cup ]-1,0[\cup ]\sqrt{2},+\infty [$