证明:(Ⅰ)∵(x3+y3)-(x2y+xy2)=x2(x-y)+y2(y-x)=(x-y)(x2-y2)=(x-y)2(x+y), 又∵x,y∈R+,∴(x-y)2≥0,,x+y>0,∴(x-y)2(x+y)≥0, ∴x3+y3≥x2y+xy2.…(5分) (Ⅱ)∵a,b,c∈R+,由(Ⅰ)知:a3+b3≥a2b+ab2;b3+c3≥b2c+bc2;c3+a3≥c2a+ca2; 将上述三式相加得:2(a3+b3+c3)≥(a2b+ab2)+(b2c+bc2)+(c2a+ca2), | 3(a3+b3+c3)≥(a3+a2b+ca2)+(b3+ab2+b2c)+(c3+bc2+c2a) | =a2(a+b+c)+b2(a+b+c)+c2(a+b+c) | =(a+b+c)+(a2+b2+c2) |
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∴a3+b3+c3≥(a2+b2+c2)(a+b+c).…(10分) |