I can't find the right way to think about the region plot of an inequality.
Considering $A=\big\{ (x,y) \in \mathbb{R^2} \mid y<x+1 \big \}$, almost automatically I say: the points "under" the line $y=x+1$ in the $(x,y)$ plane (of course the real meaning is "the points of the plane for wich the $y$ is less that the corresponding $x$, plus $1$", but in that way it is quick and easy to visualize it).
But when it comes to 3D it gets a little more confusing, for example $B=\big\{ (x,y,z) \in \mathbb{R^3} \mid x^2+y^2-2z<0 \big \}$. Here the equation $x^2+y^2-2z=0$ it's an elliptic paraboloid, and seen that "minus zero" it looks like the points of $B$ are the ones "outside" the solid, whereas the points are the ones "inside". If I rewrite the inequality as $x^2+y^2<2z$ maybe it is a little more clear, but I don't see which is the right way to visualize immediately such solid.
Firstly should I write a generic inequality in the form (for istance) $f(x,y,z)<0$ or $f(x,y)<z$ (when possible of course)? Secondly how do I interpret the two forms of the same inequalities in order to visualize clearly the solid in space? (Again the meaning of the inequalities is clear, but I'm looking for a quick and easy way to visualize it).
Thanks for your help
You're on the right track, and don't feel bad. It took me minute to see that the points are on the "inside". Start with what you call the "generic inequality". Then, if you are confused about which "side" you need to "be" on, pick any point in 3D space not on the inequality surface, and put your coordinates into the inequality. If the inequality is satisfied, you have picked the right "side". If not, you need the other "side". For instance, with $x^2 + y^2 - 2z < 0$, try $x=0, y=0, z=10$ ... definitely on the "inside", right? Is $0+0-2\times 10 < 0$? Sure is, so the "inside" is what you want.