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For all positive real numbers

  1. Jun 22, 2010 #1
    For all positive real numbers [tex]x,y[/tex] prove that:

    [tex] \frac{1}{1+\sqrt{x}}+\frac{1}{1+\sqrt{y}} \geq \frac{2\sqrt{2}}{1+\sqrt{2}}[/tex]
     
  2. jcsd
  3. Jun 22, 2010 #2
    Re: inequality

    No proof is possible.
    Let x=y=1, then left hand side is equal to 1 which is strictly less than the right hand side.
     
  4. Jun 22, 2010 #3
    Re: inequality

    I forgot to say that x,y are numbers such that x+y=1
     
  5. Jun 22, 2010 #4
    Re: inequality

    Since x+y=1, y=1-x. Now, you are looking for the minimum. How would you do it?
     
  6. Jun 22, 2010 #5
    Re: inequality

    That's why i am asking to you guys!
    This inequality is correct .. just need to be proven!
     
  7. Jun 22, 2010 #6

    Mark44

    Staff: Mentor

    Re: inequality

    Find the minimum value of
    [tex]f(x) = \frac{1}{1 + \sqrt{x} } + \frac{1}{1 + \sqrt{1 - x}}[/tex]

    This involves finding the critical points.
     
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