Question

高中数学，第16题，详细解释

Answer 1

解：
3a^3+2b^3-3a^2b-2ab^2
=3a^2(a-b)+2b^2(b-a)
=3a^2(a-b)-2b^2(a-b)
=(a-b)(3a^2-2b^2)
因为a>=b>0
故 a>b,
a^2>=b^2, 3a^2>2b^2
于是
a-b>0
3a^2-2b^2>0
于是
3a^3+2b^3-3a^2b-2ab^2>0
3a^3+2b^3>3a^2b+2ab^2

Answer 2

3a^3+2b^3-3a^2b-2ab^2=3a^3-3a^2b+2b^3-2ab^2=3a^2(a-b)+2b^2(b-a)=(a-b)(3a^2-2b^2)
因为a≥b＞0
所以a-b≥0
3a^2-2b^2＞0
所以(a-b)(3a^2-2b^2)≥0
所以
3a^3+2b^3≥3a^2b+2ab^2

Answer 3

Answer 4

即证3a*2(a-b)+2b*2(b-a)＞=0 则为(3a*2-2b*2)(a-b)>=0 ∵a>=b>0 ∴3a*2>=3b*2>2b*2 a-b>=0