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Math Help - inequality

  1. #1
    MHF Contributor red_dog's Avatar
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    inequality

    Let a, \ b>0. Prove that \left(1+\frac{a}{b}\right)^n+\left(1+\frac{b}{a}\r  ight)^n\geq 2^{n+1}, \ \forall n\in\mathbf{Z}

    I have proved for n\geq 0, but I'm stuck for n<0.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by red_dog View Post
    Let a, \ b>0. Prove that \left(1+\frac{a}{b}\right)^n+\left(1+\frac{b}{a}\r  ight)^n\geq 2^{n+1}, \ \forall n\in\mathbf{Z}

    I have proved for n\geq 0, but I'm stuck for n<0.
    Just out of curiosity, how did you prove the result for n\geqslant0. Did you use a concavity argument? If so, I think you can extend it to negative values.
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Consider f(x)=x^n and apply Jensen's inequality.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Krizalid View Post
    Consider f(x)=x^n and apply Jensen's inequality.
    Isn't that like what I said about using the concavity?
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  5. #5
    Math Engineering Student
    Krizalid's Avatar
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    Didn't see your post, sorry.

    Actually, it's even easier, we can tackle this without Jensen.

    We have \left( 1+\frac{a}{b} \right)\left( 1+\frac{b}{a} \right)=2+\frac{a}{b}+\frac{b}{a}\ge 4, hence \left( 1+\frac{a}{b} \right)^{n}+\left( 1+\frac{b}{a} \right)^{n}\ge 2\sqrt{4^{n}}=2^{n+1}.
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