Urgent!!!! P.Maths Inequality

👨🏻‍🏫 學術台數學

Felix

2009-11-11 5:57 am

x,y,z are positive real integers, show that

(xy)^2+(xz)^2+(yz)^2 >= xyz(x+y+z)

更新1:

(xy)^2+(xz)^2+(yz)^2 is equal to or greater than xyz(x+y+z)

更新2:

x,y,z are positive real NUMBERS

回答 (1)

用戶9112

2009-11-11 6:11 am

✔ 最佳答案

(xy - xz)^2 + (xz - yz)^2 + (yz - xy)^2 >= 0
(xy)^2 - 2x^2yz + (xz)^2 + (xz)^2 - 2xyz^2 + (yz)^2 + (yz)^2 - 2xy^2z + (xy)^2 >= 0
2[(xy)^2 + (xz)^2 + (yz)^2] >= 2[x^2yz + xyz^2 + xy^2z]
(xy)^2 + (xz)^2 + (yz)^2 > xyz(x + y + z)

👍 0 👎 0

收錄日期: 2021-04-23 23:23:12
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091110000051KK01697

檢視 Wayback Machine 備份