(a^a*b^b)/(ab)^[(a+b)/2]=a^[(a-b)/2]*b^[(b-a)/2]=(a/b)^[(a-b)/2]若a>b,则a/b>1,a-b>0则(a/b)^[(a-b)/2]>1^0>1则(a^a*b^b)>(ab)^[(a+b)/2]若a(ab)^[(a+b)/2]总之(a^a*b^b)>(ab)^[(a+b)/2]