(a^3+b^3)-(ba^2+ab^2)=(a^3-ba^2)-(ab^2-b^3)=(a-b)a^2-(a-b)b^2=(a^2-b^2)(a-b)=(a+b)(a-b)^2≥0a^3+b^3≥ba^2+ab^2同理可得b^3+c^3≥bc^2+cb^2,a^3+c^3≥ca^2+ac^2,三式相加得2(a^3+b^3+c^3)≥(ba^2+bc^2)+(ab^2+...