Use either a difference graph or a denominator sign test

Bring the right side to the left: $(x+1)/x-3/(x+2)\ge0$. Sketch $y=(x+1)/x-3/(x+2)=1+1/x-3/(x+2)$. It has vertical asymptotes x=−2 and x=0, horizontal asymptote y=1 and no x-intercepts. The first and third branches are positive; the middle branch is negative. Thus $\{x\in\mathbb R:x<-2\text{ or }x>0\}$. To decide this inequality, focus on the branch signs and poles; turning-point coordinates are not needed.

The left side is $[(x+1)(x+2)-3x]/[x(x+2)]=(x^2+3x+2-3x)/[x(x+2)]=(x^2+2)/[x(x+2)]$. Since x²+2>0 for every real x, require $1/[x(x+2)]\ge0$. Signs around −2,0 are +,−,+. The solution set is $\{x\in\mathbb R:x<-2\text{ or }x>0\}$. Neither pole is allowed, and equality never occurs because the numerator is positive.