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Why correct this inequality log(d/d-1)>(1/d)?

  1. Feb 19, 2014 #1
    Why correct this inequality,

    log(d/(d-1))>1/d for d≥2?
     
    Last edited: Feb 19, 2014
  2. jcsd
  3. Feb 19, 2014 #2

    maajdl

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    There must be a mistake in your question since d/d=1
    You corrected the mistake.
     
    Last edited: Feb 19, 2014
  4. Feb 19, 2014 #3
    multiplication comes before addition.
    I see d/d-1 as [tex]\frac{d}{d} -1[/tex]log 0 = who-knows-what
     
  5. Feb 19, 2014 #4

    maajdl

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    Well, check first that this inequality is correct for d=2.
    Then prove that the function log(d/(d-1)) will continue to be > than 1/d for d>2.
    You could do that by analysis the derivative.

    Get insight by making a plot of these functions.

    Alternatively, study the function log(d/(d-1))-1/d .
    Calculate its value for d=2, and analyze its behavior for d>2 .
     
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