a^3+b^3+c^3-3abc
=[( a+b)^3-3a^2b-3ab^2]+c^3-3abc
=[(a+b)^3+c^3]-(3a^2b+3ab^2+3abc)
=(a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)
=(a+b+c)(a^2+b^2+2ab-ac-bc+c^2)-3ab(a+b+c)
=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)
=（a+b+c)*1/2[(a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)]
=(a+b+c)*1/2[(a-b)^2+(a-c)^2+(b-c)^2]
>=0
題目中應該有條件是a,b,c>0吧.
用到二個公式：
a^3+b^3=(a+b)(a^2-ab+b^2)
(a+b)^3=a^3+b^3+3a^2b+3ab^2