题目

设a≥b>0,求证:3a3+2b3≥3a2b+2ab2. 答案：【证明】3a3+2b3-(3a2b+2ab2)=3a2(a-b)+2b2(b-a)=(3a2-2b2)(a-b). 因为a≥b>0,故a-b≥0,3a2-2b2>2a2-2b2=2(a+b)(a-b)≥0, 所以(3a2-2b2)(a-b)≥0,即3a3+2b3≥3a2b+2ab2.

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