Proof of Inequalities urgent
回答 (2)
2009-12-12 3:09 am
✔ 最佳答案
Since (√a - √b)^2 ≥ 0 ,therefore: a - 2√ab + b ≥ 0
a+b ≥ 2√ab
1 ≥ 2√ab / (a+b)
√ab ≥ 2ab/ (a+b )
QED
2009-12-12 11:55:58 補充：
LHS equals the RHS if and only if a = b.
2009-12-12 11:56:13 補充：
LHS equals the RHS if and only if a = b.
2009-12-22 6:37 am
LHS-RHS=(a+b)√ab-ab
=√ab(√a-√b)^2 ≥0
Therefore, √ab≥2ab/ (a+b )
LHS= RHS when a=b.
=√ab(√a-√b)^2 ≥0
Therefore, √ab≥2ab/ (a+b )
LHS= RHS when a=b.
參考: me
收錄日期: 2021-04-13 16:58:47
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