Solve $2\sqrt{(x-1)(x+2)}\ge|x+1|-2$ Can you please show me how can I solve this inequality. I would like to see how it can be done without the graph of the functions. Thank you!
$$2\sqrt{(x-1)(x+2)}\ge|x+1|-2$$

If $|x+1|-2$ is negative or zero, the inequality is obvious, so assume $|x+1| >2$, so that $x>1$ or $x < (-3)$.
If $x>1$, then $|x+1|=x+1$ so that your inequality becomes $2\sqrt{(x-1)(x+2)} \geq x-1$, which is equivalent
to $4(x-1)(x+2) \geq (x-1)^2$, or $4(x+2) \geq x-1$, i.e. $3x+9 \geq 0$ which is clear.
If $x<(-3)$, then $|x+1|=-x-1$ so that your  inequality becomes $2\sqrt{(x-1)(x+2)} \geq -x-3$, which is equivalent
to $4(x-1)(x+2) \geq (x+3)^2$, or $3x^2-2x-17 \geq 0$, which is true because
$3x^2-2x-17=16-(x+3)(11-3x)$
