设abc均为正数,且a＋b＋c＝1证明 ①a^2b^2＋b^2c^2＋c^2a^2≥abc
人气：498 ℃ 时间：2020-04-29 21:18:31
解答
abc均为正数,且a＋b＋c＝1a^2b^2＋b^2c^2＋c^2a^2-abc =a^2b^2＋b^2c^2＋c^2a^2-abc（a+b+c） =2[a^2b^2+b^2c^2+c^2a^2-abc(a+b+c)]/2=[(a^2b^2+c^2a^2-2a^2bc)+(a^2b^2+b^2c^2-2ab^2c)+(b^2c^2+c^2a^2-2abc^2)]/2 =...
推荐
- 如何证明a^4+b^4+c^4≧a^2b^2+b^2c^2+c^2a^2≥(a+b+c)abc?
- 已知abc均为正数,求证1/2a+1/2b+1/2c>1/a+b +1/b+c +1/a+c
- 已知a,b,c属于正数,求证(b^2c^2+c^2a^2+a^2b^2)/(a+b+c) ≥abc
- 已知abc是正数,求证a^2a*b^2b*c^2c大于等于a^(b+c)*b^(c+a)*c^(a+b)
- 设a,b,c是实数,满足abc=1,证明：2a-(1/b),2b-(1/c),2c-(1/a)中最多有两个数大于1
- to his surprise ,this time there were six .
- 用形容词的适当形式填空 In this country it's very ------(cold)in November,and it's much
- he stepped into the room quietly without being noticed.后面的without being noticed是什么意思?
猜你喜欢
