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Math Help - Application of AM-GM? It's not quite so simple...

  1. #1
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    Application of AM-GM? It's not quite so simple...

    Let
    a; b; c be positive reals such that abc = 1. Prove that

    {1/sqrt(b + a/2 + 1/2)} + {1/sqrt(c + b/2 + 1/2)} + {1/sqrt(a + c/2 + 1/2)} >= sqrt(2)

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  2. #2
    MHF Contributor
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    Advanced Algebra?
    I only studied retarded/subdued/garden-variety Algebra. So I'll attack your question here with my crude tools, and then I'll leave it hanging for others who were/are into Advanced Algebra.

    {1/sqrt(b + a/2 + 1/2)} + {1/sqrt(c + b/2 + 1/2)} + {1/sqrt(a + c/2 + 1/2)} >= sqrt(2)

    {1/sqrt[(2b+a+1)/2]} + {1/sqrt[(2c+b+1)/2]} + {1/sqrt[(2a+c+1)/2]} >= sqrt(2)

    {sqrt(2)/sqrt(2b+a+1)} + {sqrt(2)/sqrt(2c+b+1)} + {sqrt(2)/sqrt(2a+c+1)} >= sqrt(2)

    {1/sqrt(2b+a+1)} + {1/sqrt(2c+b+1)} + {1/sqrt(2a+c+1)} >= 1

    I'm up to there only.
    Maybe I'd expand that by eliminating all the radical signs, but I'm not yet too crazy.

    {1/sqrt(2b+a+1)} + {1/sqrt(2c+b+1)} + {1/sqrt(2a+c+1)} >= abc ?
    I don't know that too.
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  3. #3
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    Wait, I'm sorry I didn't post the problem correctly. The problem is actually

    Prove that

    {1/sqrt(b + 1/a + 1/2)} + {1/sqrt(c + 1/b + 1/2)} + {1/sqrt(a + 1/c + 1/2)} >= sqrt(2)



    Once again, I apologize.
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  4. #4
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    Florida
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    well for starters I advise that you don't begin with AM-GM. instead use AM-HM
    which says that for positive \alpha, \beta, \gamma that \frac{\alpha+\beta+\gamma}{3}\geq \frac{3}{\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}  {\gamma}}

    so using this we that the expression on the left (with out the square roots) is greater than or equal to

    \frac{9}{a+b+c+\frac{\alpha}{2}+\frac{\beta}{2}+\f  rac{\gamma}{2}+\frac{3}{2}} which is less than

    \frac{9}{a+b+c+\frac{\alpha}{2}+\frac{\beta}{2}+\f  rac{\gamma}{2}}

    NOW from AM-GM we have that this is less than \frac{9}{3+\frac{3}{2}}=2

    now that you know this you can do a little modification to get \sqrt{2}
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