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Prove ln(x) < sqrt(x) for all x>0

  1. Sep 28, 2010 #1

    I wonder how to prove that ln(x) < sqrt(x) for all x>0?

    Please enlighten me on two possible way to prove this .
    Proof1. Using calculus and derivatives
    Proof2. Since I'm taking real analysis, I wonder if it is possible to use taylor series to show this in an elegant way.
  2. jcsd
  3. Sep 28, 2010 #2
    I don't know a "proof2", here's a "proof1" (idea, you do the calculations!):

    1) Prooving your statement is equivalent to prooving that f(x) = ln(x) - sqrt(x) < 0 for all x > 0.

    2) Calculate f ' (x).

    3) Find y such that f ' (y) = 0.

    4) Show that f (y) < 0.

    5) Show that the limits for x -> 0+ and x -> +infty of f(x) are negative.

    6) Why points 1) - 5) proove your claim?
  4. Sep 28, 2010 #3


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