1,(ab+a+b+1)(ab+ac+bc+c^2)=(a+1)(b+1)(a+c)(b+c)≥2√a*2√b*2√ac*2√bc=16abc
2,a^3+b^3-a^2*b-b^2*a=(a-b)^2(a+b)≥0,所以a^3+b^3≥a^2*b+b^2*a
同理b^3+c^3≥b^2*c+c^2*b,a^3+c^3≥a^2*c+c^2*a
左边与左边相加,右边与右边相加,整理即可得到
1,(ab+a+b+1)(ab+ac+bc+c^2)=(a+1)(b+1)(a+c)(b+c)≥2√a*2√b*2√ac*2√bc=16abc
2,a^3+b^3-a^2*b-b^2*a=(a-b)^2(a+b)≥0,所以a^3+b^3≥a^2*b+b^2*a
同理b^3+c^3≥b^2*c+c^2*b,a^3+c^3≥a^2*c+c^2*a
左边与左边相加,右边与右边相加,整理即可得到