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Proving Bernoulli's Inequality

  1. Jul 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove Bernoulli's Inequality: if ##h>-1##
    [itex](1+h)^n \geq 1+hn[/itex]

    2. Relevant equations
    Binomial Theorem
    [itex](a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}[/itex]

    3. The attempt at a solution
    If ##h=0##
    [itex](1+0)^n=1[/itex]
    [itex]1=1[/itex]

    If ##h>0##
    This
    [itex](1+h)^n \geq 1+hn[/itex]
    Implies
    [itex](1+h)^n=\sum_{k}^{n}\binom{n}{k}h^{k}[/itex]
    [itex]\sum_{k=2}^{n}\binom{n}{k}h^{k} \geq 0[/itex]

    So the proof is done.
     
  2. jcsd
  3. Jul 22, 2013 #2

    micromass

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    And what if ##h## is negative?
     
  4. Jul 22, 2013 #3
    Owh, I keep forgetting that the number after -1 is not 0.

    ##h>-1## is equivalent to ##h+1>0##

    We know [itex]\sum_{k=2}^{n}\binom{n}{k}h^{k} \geq 0[/itex]

    Under the closure of multiplication positive number will always give positive number so:
    [itex](h+1)\sum_{k=2}^{n}\binom{n}{k}h^{k} \geq 0[/itex]

    [tex]\sum_{k=2}^{n}\binom{n}{k}h^{k+1}+\sum_{k=2}^{n}\binom{n}{k}h^{k}\geq 0[/tex]

    Is the reasoning fine?
     
  5. Jul 22, 2013 #4

    micromass

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    I am not convinced of this if ##h## is negative. It requires a proof.
     
  6. Jul 22, 2013 #5
    Ok will proof by induction works in this case, if there's negative number in play?
     
  7. Jul 22, 2013 #6

    micromass

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    Induction on ##n## will work. Not sure why you think ##h## being negative will make an induction proof invalid.
     
  8. Jul 22, 2013 #7
    Ahh, I didn't see that. I'm just a bit confused perhaps.

    So the plan:
    1.Proof the inequality is true using n=1
    2.Assume n is true, to show n+1 is also true.

    [itex](1+h)^{n+1}=(1+h)(1+h)^n[/itex]
    [itex](1+h)(1+h)^n=(1+h)(1+nh)[/itex]
    [itex](1+h)(1+nh)=1+nh+h+nh^2=1+(n+1)h+nh^2[/itex]
    By inspection this should be true ##1+(n+1)h+nh^2 \gek 1+(n+1)h## for any h, and so the Bernoulli's inequality in consequence.
     
    Last edited: Jul 22, 2013
  9. Jul 22, 2013 #8

    micromass

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    This equality isn't true.

     
  10. Jul 22, 2013 #9
    Ok I think I know where that goes wrong.

    [itex](1+h)^{n+1}=(1+h)(1+h)^n[/itex]
    [itex](1+h)(1+h)^n \geq (1+h)(1+nh)[/itex] assuming the equality holds with ##n##
    [itex](1+h)(1+nh)=1+nh+h+nh^2=1+(n+1)h+nh^2[/itex]
    ##1+(n+1)h+nh^2 \geq 1+(n+1)h##
    then
    [itex](1+h)^{n+1} \geq 1+(n+1)h[/itex]

    This is true right?
    ##1+(n+1)h+nh^2 \geq 1+(n+1)h##
    ##nh^2 \geq 0##
     
  11. Jul 22, 2013 #10

    micromass

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    OK, that seems right.
     
  12. Jul 22, 2013 #11
    Finally! Thank you for guiding me and clearing all of my messy mistakes. :D

    Next question.
     
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