(b) Show that [(a+b+c+d)/4]^4 >= abcd
(c) Using (b), show that (a+b+c)/3]^3 >= abc
更新1:
If a,b,c are positive numbers. thankyou!
A.maths inequality
2013-09-12 5:34 am
(a) Show that [(a+b)/2]^2 >= ab
(b) Show that [(a+b+c+d)/4]^4 >= abcd
(c) Using (b), show that (a+b+c)/3]^3 >= abc
(b) Show that [(a+b+c+d)/4]^4 >= abcd
(c) Using (b), show that (a+b+c)/3]^3 >= abc
回答 (2)
2013-09-12 6:50 am
✔ 最佳答案
典型 Pure Math 題。圖片參考：http://imgcld.yimg.com/8/n/HA00430218/o/20130911224955.jpg
2013-09-11 23:16:30 補充：
對對對！
很有道理～
相信題目應該要說清是 non-negative numbers.
2013-09-12 6:05 am
a , b , c ≥ 0 ??
2013-09-11 23:09:32 補充：
c) 部也可令 d = (a+b+c)/3 , 則
[( a + b + c + (a+b+c)/3 ) / 4]⁴≥ abc(a+b+c)/3
[(a + b + c) / 3]⁴≥ abc(a+b+c)/3
[(a + b + c) / 3]³ ≥ abc
2013-09-11 23:09:32 補充：
c) 部也可令 d = (a+b+c)/3 , 則
[( a + b + c + (a+b+c)/3 ) / 4]⁴≥ abc(a+b+c)/3
[(a + b + c) / 3]⁴≥ abc(a+b+c)/3
[(a + b + c) / 3]³ ≥ abc
收錄日期: 2021-04-21 22:26:38
原文連結 [永久失效]:
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