a>0 b>0(a∧a)×(b∧b)/[(ab)∧(a/2＋b/2)]=(a/b)^(a/2)/(b/a)^(b/2)=(a/b)^[(a-b)/2](1) a>b a/b>1 a-b>0所以(a/b)^[(a-b)/2]>1(2)a=b (a/b)^[(a-b)/2]=1(3)a