Note that a^2 >= 0 for any real number a. If a and b are two real numbers, prove that
a) a^2+b^2 >= 2ab
b) (a+b)/s >= (ab)^(1/2)
how to link a^2 >= 0 to the inequality to prove it?
FORM 3 MATHS
2014-10-18 10:32 pm
回答 (3)
2014-10-19 5:58 am
✔ 最佳答案
a) (a-b)^2 > 0 <since a and b are real number, square of real number must > 0
a^2+b^2-2ab > 0
so, a^2+b^2 > 2ab
b)
I don't know where is "s" come from, i can only tell you that
(ab)^(1/2) = root ab
2014-10-19 14:16:10 補充：
square of real number should be >=0, which is non-negative number, you will learn that next year
2014-10-20 12:55 am
我被人檢舉了。
不知道又得罪了誰?
不知道又得罪了誰?
2014-10-18 11:25 pm
a)
(a - b) is a real number.
(a - b)^2 ≥ 0
a^2 + b^2 - 2ab ≥ 0
a^2 + b^2 - 2ab + 2ab ≥ 0 + 2ab
a^2 + b^2 ≥ 2ab
b)
[a^(1/2) - b^(1/2)] is a real number.
[a^(1/2) - b^(1/2)]^2 ≥ 0
[a^(1/2)]^2 + [b^(1/2)]^2 - 2[a^(1/2)][b^(1/2)] ≥ 0
a + b - 2(ab)^(1/2) ≥ 0
a + b ≥ 2(ab)^(1/2)
(a + b)/2 ≥ (ab)^(1/2)
(a - b) is a real number.
(a - b)^2 ≥ 0
a^2 + b^2 - 2ab ≥ 0
a^2 + b^2 - 2ab + 2ab ≥ 0 + 2ab
a^2 + b^2 ≥ 2ab
b)
[a^(1/2) - b^(1/2)] is a real number.
[a^(1/2) - b^(1/2)]^2 ≥ 0
[a^(1/2)]^2 + [b^(1/2)]^2 - 2[a^(1/2)][b^(1/2)] ≥ 0
a + b - 2(ab)^(1/2) ≥ 0
a + b ≥ 2(ab)^(1/2)
(a + b)/2 ≥ (ab)^(1/2)
收錄日期: 2021-04-16 16:38:05
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