y'=( 1-lnx)/x^2,a>； 0。 Yes: f' (x) = a (1-lnx)/x 2, when 0; 0，F '(x)& gt； 0, F(x) increases, and the minimum value is removed when x=2a and F(a)=a*(lna)/a=lna.

When a<= e< =2a, F(x) first increases and then decreases. Find the value of F(x) at point A and point 2A respectively, and then compare the sizes.

When e> is at 2a, F(x) decreases monotonically, and the minimum value of F(x) is F(2a)=a*(ln2a)/2a=(ln2a)/2.