I'm tempted to separate this inequality into two parts:
x > 0 and 0 < y + z.
x > 0 includes all of the 3-dimensional solid to the right of the plane x = 0.
If 0 < y + z, then -y < z, or z > -y. z = -y is the equation of a plane that is good for all x (x doesn't matter here). Draw the plane (thru the origin) that satisfies z = -y. Then 0 < y + z denotes the portion of the 3-d solid BELOW the plane z = -y.
Best to draw this in 3-d and to shade the portions of the solid centered at (0, 0, 0) that satisfy both x > 0 and 0 < y + z.