解析：∵a^3+b^3+c^3-3abc
=(a+b)^3-3ab（a+b)+c^3-3abc
=(a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)
=(a+b+c)[(a+b)^2-(a+b)c+c^2-3ab]
=(a+b+c）（a^2+b^2+c^2-ab-bc-ac)
=1/2*(a+b+c)(2a^2+2b^2+2c^2-2ab-2bc-2ac)
=1/2*(a+b+c)[(a-b)^2+(b-c)^2+(a-c)^2]
∵a＞0,b＞0,c＞0
∴a+b+c＞0
(a-b)^2≥0,(b-c)^2≥0,(a-c)^2≥0,
則1/2*(a+b+c)[(a-b)^2+(b-c)^2+(a-c)^2]≥0
即a^3+b^3+c^3-3abc≥0
∴(a^3+b^3+c^3)/3≥abc
那么(a+b+c)/3≥(abc)^(1/3)
a,b,c是屬于正實(shí)數(shù)成立,不是你說的正有理數(shù).