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Math Help - prove

  1. #1
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    prove



    i have a problem to prove this..

    Let a,b,c be positive real numbers such that abc=1, prove that

    1/(a^3*(b+c))+1/(b^3*(c+a))+1/(c^3*(a+b))>=3/2.

    can anyone solve this problem? thank you

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  2. #2
    MHF Contributor red_dog's Avatar
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    Let a=\frac{1}{x}, \ b=\frac{1}{y}, \ c=\frac{1}{z}\Rightarrow xyz=1

    \frac{1}{a^3(b+c)}=\frac{1}{\frac{1}{x^3}\left(\fr  ac{1}{y}+\frac{1}{z}\right)}=\frac{x^3yz}{y+z}=\fr  ac{x^2}{y+z}

    The inequality becomes

    \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge  q\frac{3}{2}

    By Cauchy-Schwarz

    [(y+z)+(z+x)+(x+y)]\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x  +y}\right)\geq(x+y+z)^2

    or \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge  q\frac{x+y+z}{2}\geq\frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Since abc=1\implies(abc)^2=1, both sides positives since the assumption a,b,c\in\mathbb R^+ is given, so the LHS is \frac{a^{2}b^{2}c^{2}}{a^{3}(b+c)}+\frac{a^{2}b^{2  }c^{2}}{b^{3}(a+c)}+\frac{a^{2}b^{2}c^{2}}{c^{3}(a  +b)}=\frac{(bc)^{2}}{a(b+c)}+\frac{(ac)^{2}}{b(a+c  )}+\frac{(ab)^{2}}{c(a+b)}. Now apply Cauchy–Schwarz and get \frac{(bc)^{2}}{a(b+c)}+\frac{(ac)^{2}}{b(a+c)}+\f  rac{(ab)^{2}}{c(a+b)}\ge \frac{(ab+bc+ac)^{2}}{2(ab+bc+ac)}=\frac{ab+bc+ac}  {2}. Hence since ab+bc+ac\ge 3\sqrt[3]{(abc)^2}=3, we have our inequality.
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