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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

    micromass

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    Hi ArcanaNoir! :smile:

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

    [tex]a+\frac{b}{n}=3[/tex]

    but

    [tex]a\sqrt{n}{b}=2\sqrt{2}[/tex]

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

    Looking at the graph of

    [tex]x+\frac{y}{2}-x\sqrt{y}[/tex]

    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]
    AND
    [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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