a≥b﹥0,求证:3a³+2b³≥3a²b+2ab²

a≥b﹥0,求证:3a³+2b³≥3a²b+2ab²
a≥b﹥0,求证:3a³+2b³≥3a²b+2ab²

a≥b﹥0,求证:3a³+2b³≥3a²b+2ab²
3a³+2b³-(3a²b+2ab²)
=3a²(a-b)+2b²(b-a)
=3a²(a-b)-2b²(a-b)
=(a-b)(3a²-2b²)
=(a-b)[2(a²-b²)+a²]
∵a≥b>0
∴a-b≥0,
a²≥b²
∴(a-b)[2(a²-b²)+a²]≥0
∴3a³+2b³≥3a²b+2ab²