a>b>c>0, ∴a/b>1,(a-b)/3>0, ∴(a/b)^[(a-b)/3]>1 ……(1) 同理,有 (a/c)^[(a-c)/3]>1 ……(2) (b/c)^[(b-c)/3]>1 ……(3) 由(1)×(2)×(3),得 (a/b)^(a-b)/3*(b/c)^(b-c)/3*(a/c)^(a-c)/3>1 →a^[(a-b)/3+(a-c)/3]*b^[(b-a)/3+(b-c)/3]*c^[(c-a)/3+(c-b)/3]>1 →(a^a*b^b*c^c)/(abc)^[(a+b+c)/3]>1 →a^a*b^b*c^c>(abc)^[(a+b+c)/3].
答：用比商法证明，如下图所示（点击放大图片）详情>>
答：详情>>
