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设αβγ均为锐角设α、β、γ均为锐角,且(sinα)^2+(si

2011-03-29 21:30:08s***
设α、β、γ均为锐角,且(sinα)^2+(sinβ)^2+(sinγ)^2=1,求证:(sinα)^3/sinβ+(sinβ)^3/sinγ+(sinγ)^3/sinα≥1.设αβγ均为锐角设α、β、γ均为锐角,且(sinα)^2+(sinβ)^2+(sinγ)^2=1,求证:(sinα)^3/sinβ+(sinβ)^3/sinγ+?

最佳回答

  • 更为简洁的是：令a=sinα,b=sinβ,c=sinγ a^3/b+ab≥2a^2,b^3/c+bc≥2b^2,c^3/a+ab≥2c^2 三式相加得 a^3/b+b^3/c+c^3/a+ab+bc+ca≥2(a^2+b^2+c^2) 而ab+bc+ca≤a^2+b^2+c^2 所以a^3/b+b^3/c+c^3/a≥a^2+b^2+c^2=1 当且仅当a=b=c时取等号 原不等式获证！
    2011-03-30 08:08:34
  • 证明:由三元均值不等式,得 (sinα)^3/sinβ+(sinα)^3/sinβ+sin^2β≥3sin^2α, (sinβ)^3/sinγ+(sinβ)^3/sinγ+sin^2γ≥3sin^2β, (sinγ)^3/sinα+(sinγ)^3/sinα+sin^2α≥3sin^2γ,三式相加,得 [(sinα)^3/sinβ+(sinβ)^3/sinγ+(sinγ)^3/sinα]+ [(sinα)^2+(sinβ)^2+(sinγ)^2]≥3[(sinα)^2+(sinβ)^2+(sinγ)^2],将(sinα)^2+(sinβ)^2+(sinγ)^2=1代入,得 (sinα)^3/sinβ+(sinβ)^3/sinγ+(sinγ)^3/sinα≥1.
    2011-03-29 22:15:09
  • α、β、γ均为锐角,即sinα、sinβ、sinγ>0, 为方便手机表达,设sinα=x,sinβ=y,sinγ=z, 则依柯西不等式得, (xy+yz+zx)(x^3/y+y^3/z+z^3/x)≥(x^2+y^2+z^2)^2 →x^3/y+y^3/z+z^3/x≥1/(xy+yz+zx)……(1) 而(x^2+y^2+z^2)(y^2+z^2+x^2)≥(xy+yz+zx)^2(柯西不等式) →xy+yz+zx≤1 →1/(xy+yz+zx)≥1 ……(2) 以(2)代入(1)得 x^3/y+y^3/z+z^3/x≥1. 即(sinα)^3/sinβ+(sinβ)^3/sinγ+(sinγ)^3/sinα≥1.
    2011-03-29 22:09:48
    • 很赞哦！ (124)

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