It's a matter of taking moments about the reference point to determine the extra tail load.
If the cg is X forward of the reference point and the tailplane is L aft of the reference point, then the extra download on the tail, T, due to the CG being forward is:
T = W * (X/L)
So the total "effective weight" is the actual weight plus the extra download, i.e. W+T
So that means that when CLmax was deteremined with the CG at the reference, for a weight W0, now that lift has to support W+T
W0 = W+T = W + W * (X/L) = W * { 1 + (X/L) }
reversing the equation, we get
W = W0 / { 1 + (X/L) }
But our reference CLmax0 came from the W0 case, so our new CLmax would be reduced by the same factor, i.e.
CLmax = CLmax0 / { 1 + (X/L) }
So the further forward the CG moves, the more CLmax appears to reduce. the longer the tail arm is, the less a given CG shift affects it.
Note that X and L both have to be in the same units - if L is in metres, so is X - you can't just use the %CG shift, you need to multiply it by the reference dimension (the mean chord, usually). Also, typically the reference CG is 25% of the chord, and the reference tail arm measured from the 25% wing chord position to the 25% tail chord position.