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Is this equation provable?

  1. Jul 26, 2011 #1
    1. The problem statement, all variables and given/known data
    I'm working on a Cauchy inequality proof, and I'm at a point where I need to prove:
    [itex] a+ \frac{b}{n} = a\sqrt[n]{b} [/itex] iff [itex] a=b [/itex]

    Anyway, is it even possible to prove this? If it isn't I need to hurry up and get about another method. If it is possible, I don't mind thinking on it some more.
    Last edited: Jul 26, 2011
  2. jcsd
  3. Jul 26, 2011 #2
    Hi ArcanaNoir! :smile:

    This doesn't seem true. Take a=b=n=2, then




    and these are certainly not equal. So at least one of the implications fails.

    Looking at the graph of


    suggest that the other implication fails as well.
  4. Jul 27, 2011 #3
    Oh dear.
  5. Jul 27, 2011 #4
    Okay, I think I fixed it.
    [tex] \frac{na+b}{n+1}=\sqrt[n+1]{a^nb} [/tex]
  6. Jul 27, 2011 #5
    I'm not well versed in proof writing but couldn't you just algebraically show it's true for a=b?

    As in:

    [tex] \frac{na + a}{n+1} = \sqrt[n+1]{a^na} [/tex]
    [tex] \frac{(n+1)a}{n+1} = \sqrt[n+1]{a^{n+1}} [/tex]
    [tex] a = a [/tex]

    Though this would require that n cannot equal -1, but I think you may already have that restriction.
  7. Jul 27, 2011 #6
    This would only prove one implication. This is an if and only if proof, so you have to prove that
    [tex]a=b \implies \frac{na + b}{n+1}=\sqrt[n+1]{a^nb} [/tex]
    [tex]\frac{na + b}{n+1} = \sqrt[n+1]{a^nb} \implies a=b[/tex]
    Last edited: Jul 27, 2011
  8. Jul 27, 2011 #7
    isn't that what the op wants though? This statement is true iff a=b...
  9. Jul 27, 2011 #8
    BrianMath is right. It's the second thing I need most, the first part of the biconditional is easy.

    A and b are greater than zero and n is a natural number.
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