若a.b.c为正整数,求a3+b3+c3与3abc的大小.
2个回答

证明:a^3+b^3+c^3

=(a+b)(a^2-ab+b^2)+c^3

=(a+b)^3-3ab(a+b)+c^3

=(a+b+c)^3-3c(a+b)(a+b+c)-3ab(a+b)

=(a+b+c)^3-3c(a+b)(a+b+c)-3ab(a+b+c)+3abc

=(a+b+c)[(a+b+c)^2-3c(a+b)-3ab]+3abc

=(a+b+c)(a^2+b^2+c^2+2ab+2bc+2ac-3ac-3bc-3ab)+3abc

=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)+3abc

=0.5(a+b+c)(2a^2+2b^2+2c^2-2ab-2ac-2bc)+3abc

=0.5(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]+3abc≥3abc

显然当且仅当a=b=c时等号成立.