設0
2007-12-04
18:00:24
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4 個解答
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發問者
倫倫
1
in
科學
➔ 數學
pf: 因為(a-1)(b-1)(c-1)=a b c-ab-bc-ca-1 abc<=0--------(1)
(ab-1)(bc-1)(ca-1)=a^2*b^2*c^2 ab bc ca-(a b c)*abc-1<=0----(2)
(1) (2)得 a b c-2 abc a^2*b^2*c^2-(a b c)*abc<=0
(abc-1)(abc-a-b-c 2)<=0
因為abc-1<=0 所以 abc-a-b-c 2>=0 , a b c-2-abc<=0-----(3)
原左式=c/(ab 1) b/(ca 1) a/(bc 1)<=(a b c)/(abc 1)
又(a b c)/(abc 1)-2=(a b c-2-2abc)/(abc 1)
由(3)得上式必<=0
故 c/(ab 1) b/(ca 1) a/(bc 1)<=(a b c)/(abc 1)<=2 得証
2007-12-07 10:13:19 補充：
抱歉,我不太懂為何打上去的"+"號會消失!
2007-12-06 10:22:57 · answer #1 · answered by 小潘潘 1 · 0⤊ 0⤋
解得很漂亮喔!可是"+"號都不見了!
2007-12-06 20:15:05 · answer #2 · answered by mathmanliu 7 · 0⤊ 0⤋
羊大大，你二點多了還在線上哦！不睡覺嗎？
2007-12-05 04:24:26 · answer #3 · answered by tsl 7 · 0⤊ 0⤋
這之前有人證明過了 稍微找一下吧
2007-12-07 12:13:37 補充：
回覆二樓的
唸書無聊了就上來逛逛
2007-12-04 21:06:45 · answer #4 · answered by 失去羽翼的羊 6 · 0⤊ 0⤋