a^ab^b/{(ab)^[(a+b)/2]}=a^[(a-b)/2]*b^[(b-a)/2]=(a/b)^[(a-b)/2]∵ a>b>0∴ a/b>1 a-b>0则 (a/b)^[(a-b)/2]>=(a/b)^0=1故 (a/b)^[(a-b)/2]>1即 a^ab^b>(ab)^[(a+b)/2]