problem in math?
回答 (2)
2016-01-27 1:30 pm
For a,b ≥ 0, by A.M. ≥ G.M. :
a³ + b³ = a³/2 + a³/2 + b³ ≥ 3∛(a³/2 ‧a³/2 ‧b³) = 3a²b/∛4
⇔ 3a²b ≤ ∛4 (a³ + b³)
The equality holds if and only if a³/2 = a³/2 = b³ ⇔ a = ∛2 b.
a³ + b³ = a³/2 + a³/2 + b³ ≥ 3∛(a³/2 ‧a³/2 ‧b³) = 3a²b/∛4
⇔ 3a²b ≤ ∛4 (a³ + b³)
The equality holds if and only if a³/2 = a³/2 = b³ ⇔ a = ∛2 b.
2016-01-27 4:40 pm
3a²b/∛4 = 3∛(a³/2 ‧a³/2 ‧b³) ≤ a³/2 + a³/2 + b³ = a³ + b³
3a²b ≤ ∛4 (a³ + b³)
3a²b ≤ ∛4 (a³ + b³)
收錄日期: 2021-04-18 14:21:18
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160127044804AAr14f8