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题 :已知ａ、ｂ、ｃ∈Ｒ+且ａ +ｂ+ｃ =1求证 (ａ+1ａ) (ｂ+1ｂ) (ｃ+1ｃ) ≥1 0 0 02 7①分析 证明此题的关键在三个方面 :(1 )等号何时成立 :(2 )怎样拆项 ;(3 )会用平均值不等式 .易知ａ=ｂ =ｃ=13 时①取等号 ,①等价于 (3ａ+3ａ) (3ｂ+3ｂ) (3ｃ+3ｃ) ≥ 1 0 0 0 .将 3
Title: Known a, b, c∈R+ and a +b+c =1 Proof (a+1a) (b+1b) (c+1c) ≥1 0 0 02 71 The analysis proves that the key of this question is in three. Aspects: (1) When the equal sign is established: (2) How to remove the item; (3) Use the average inequality. Easy to know a=b = c=13 When 1 takes the equal sign, 1 is equivalent to (3a+ 3a) (3b+3b) (3c+3c) ≥ 1 0 0 0. Will 3