Okay, I had asignerror of all things!

The simplest way to handle a complicated inequality is to first solve the associatedequation.

That is, look at 2m+ 2n- mn= 3 or, preferably, xy- 2x- 2y+ 3= 0.

If we let x= x'- a and y= y'- b then xy- 2x- 2y+ 3= x'y'- ax'- ay'+ab- 2x'+ 2a- 2y'+ 2b+ 3= x'y'- (a+2)x'- (b+2)y'+ (3+ 2a+ 2b+ ab)= 0. In particular, taking a= b= -2 reduces that to x'y'+ 3- 4= x'y' - 1= 0 or x'y'= 1. That is a hyperbola with asymptotes parallel to the x and y axes and center at (0, 0) in the x'y'-system or (2, 2) in the xy-system.

The point is that the points where that isequalto 3 separate "< 3" from "> 3". Since (2, 2) is "between" the two branches and 2(2)+ 2(2)- (2)(2)= 4> 3 it follows that 2x+ 2y- xy> 3 forall(x, y) between the two branches of the parabola. On the other hand, (8, 8) is outside one branch of the hyperbola and 2(8)+ 2(8)- (8)(8)= 32- 64= -32< 3 and (-8, -8) is outside the other branch and 2(-8)- 2(-8)- (-8)(-8)= -32- 64= -96< 3 so points between the two branches of the hyperbola are the only (x, y) that satisfy the inequality.