Results 1 to 5 of 5

Thread: inequality

  1. #1
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    22
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    15
    Consider f(x)=x^n and apply Jensen's inequality.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    22
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    15
    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}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Jan 11th 2011, 09:20 PM
  2. Replies: 3
    Last Post: Dec 12th 2010, 02:16 PM
  3. inequality
    Posted in the Math Challenge Problems Forum
    Replies: 7
    Last Post: Jul 25th 2010, 07:11 PM
  4. Inequality help
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: Jul 8th 2010, 07:24 AM
  5. Inequality :\
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Oct 12th 2009, 02:57 PM

Search Tags


/mathhelpforum @mathhelpforum