设a≥b>0,求证:3a3+2b3≥3a2b+2ab2.
问题描述:
设a≥b>0,求证:3a3+2b3≥3a2b+2ab2.
答
证明:(3a3+2b3)-(3a2b+2ab2)=3a2(a-b)+2b2(b-a)
=(a-b)(3a2-2b2)
∵a≥0,b>0,∴a-b≥0,3a2-2b2>0
∴(3a3+2b3)-(3a2b+2ab2)≥0
∴3a3+2b3≥3a2b+2ab2.