不等式左边移到右边,有:
(a^2b+b^2c+c^2a)-(ab^2+bc^2+ca^2)
(a^2b+b^2c+c^2a)-(ab^2+bc^2+ca^2)
=ab(a-b)+c(b^2-a^2)+c^2(a-b)
=(a-b)(ab-c(a+b)+c^2)
=(a-b)[a(b-c)-c(b-c)]
=-(a-b)(b-c)(c-a)>0
所以成立
设a<b<c,求证bc^2+ca^2+ab^2< b^2c+c^2a+a^2b 11
设a<b<c,求证bc^2+ca^2+ab^2< b^2c+c^2a+a^2b 11
数学人气:457 ℃时间:2019-12-06 18:01:45
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