a /b^2 + b/a^2 ≥ 1/a + 1/b
不等式的證明1
2011-03-30 6:28 am
回答 (2)
2011-03-30 6:38 am
2011-03-30 6:41 am
(a/b^2)+(b/a^2)>=(1/a)+(1/b)
L.H.S=(ab^2+a^2b)/(a^2b^2)
=ab(a+b)/(a+b)^2
=(a+b)/ab
R.H.S=(b/ab)+ (a/ab)
=(a+b)/ab
Because L.H.S>=R.H.S,Therefore
a /b^2 + b/a^2 ≥ 1/a + 1/b is satisfied
L.H.S=(ab^2+a^2b)/(a^2b^2)
=ab(a+b)/(a+b)^2
=(a+b)/ab
R.H.S=(b/ab)+ (a/ab)
=(a+b)/ab
Because L.H.S>=R.H.S,Therefore
a /b^2 + b/a^2 ≥ 1/a + 1/b is satisfied
參考: me
收錄日期: 2021-04-23 23:23:29
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110329000051KK01476