(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
Urgent!!!! P.Maths Inequality
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)
(xy)^2+(xz)^2+(yz)^2 >= xyz(x+y+z)
回答 (1)
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)
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