不等式的証明

2011-03-24 5:12 am
設a.b.c為三角形ABC中的三條邊長,求証1/(b+c-a) +1/(c+a-b) +1/(a+b-c)>=a/(a+b+c)
更新1:

後面係9/(a+b+c)

回答 (1)

2011-03-24 6:16 am
✔ 最佳答案
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2011-03-27 21:47:45 補充:
(a+b+c)/3 >= (abc)^(1/3) (this is AM >= GM)
(1/a+1/b+1/c)/3 >= (1/abc)^(1/3) (this is also AM >= GM)
=> (abc)^(1/3) >= 3/(1/a+1/b+1/c) (this is called harmonic mean HM)
So (a+b+c)/3 >= 3/(1/a+1/b+1/c) (the result is AM >= HM)


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