THX~
更新1:
看不明，何謂AM>GM
不等式的証明
回答 (2)
2007-02-04 5:53 am
✔ 最佳答案
using AM>=GM(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)/6>=[(a^2b)(a^2c)(b^2a)(b^2c)(c^2a)(c^2b)]^1/6
(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)/6>=(a^6b^6c^6)^1/6
(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)/6>=abc
(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)>=6abc
2007-02-03 23:02:23 補充：
http://home.netvigator.com/~leeleung/theorem_0004.htmlhere
2007-02-04 5:55 am
這是 AM >= GM 的應用.
0.5 ( a + b ) >= (ab)^0.5
0.5 ( b + c ) >= (bc)^0.5
0.5 ( a + c ) >= (ac)^0.5
therefore 0.125 ( a + b ) ( b + c ) ( b + c ) > = (ab*bc*ca)^ 0.5
( a + b ) ( b + c ) ( b + c ) > = 8 abc
and ( a + b ) ( b + c ) ( b + c ) = a^2b+a^2c+b^2a+b^2c+c^2a+c^2b + 2abc
so a^2b+a^2c+b^2a+b^2c+c^2a+c^2b>6abc
0.5 ( a + b ) >= (ab)^0.5
0.5 ( b + c ) >= (bc)^0.5
0.5 ( a + c ) >= (ac)^0.5
therefore 0.125 ( a + b ) ( b + c ) ( b + c ) > = (ab*bc*ca)^ 0.5
( a + b ) ( b + c ) ( b + c ) > = 8 abc
and ( a + b ) ( b + c ) ( b + c ) = a^2b+a^2c+b^2a+b^2c+c^2a+c^2b + 2abc
so a^2b+a^2c+b^2a+b^2c+c^2a+c^2b>6abc
收錄日期: 2021-04-25 16:52:13
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070203000051KK04409