我有條題目唔知點做..可唔可以搵人教下我
let a,b,c be positive unequal real number
using the results a^2 + b^2 > 2 ab ,, b^2 + c^2 > 2bc
and c^2 + a^2 > 2 ac
1) deduce that a^2 - ab + b^2 >ab
2) deduce that a^2 + b^2 + c^2 > bc + ca + ab
3) show that a^3 + b^3 > ab( a+b)
thz! =)
amaths
2009-10-04 10:40 am
回答 (1)
2009-10-04 7:34 pm
✔ 最佳答案
1. By a2 + b2 > 2abAs a and b are positive numbers,
So, ab is also a positive number.
Therefore, a2 + b2 - ab > 2ab - ab
a2 - ab + b2 > ab
2. a2 + b2 > 2ab
b2 + c2 > 2bc
c2 + a2 > 2ac
As all numbers are positive, we can simply add up the inequalities,
2(a2 + b2 + c2) > 2(bc + ca + ab)
a2 + b2 + c2 > bc + ca + ab
3. a3 + b3 = (a + b)(a2 - ab + b2)
> ab(a + b) (By part a)
參考: Physics king
收錄日期: 2021-04-19 15:27:29
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091004000051KK02150