Geometric and Arithmetic Means Watch

RMorley
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#1
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could anyone point me on a direction to approach the following question please:

"Prove that the arithmetic means of any two different positives numbers exceeds the geometric mean of the same two numbers"

Many thanks for your help
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SimonM
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Write out what it means: \displaystyle \frac{a+b}{2} \geq \sqrt{ab}

See if you can rearrange this to make it clear why it's true
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Kolya
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One method for proving the problem is to use proof by contradiction. That is: Assume the opposite of your statement, and show that a contradiction follows.
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RMorley
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(Original post by SimonM)
Write out what it means: \displaystyle \frac{a+b}{2} \geq \sqrt{ab}

See if you can rearrange this to make it clear why it's true
Thanks, in the end I get to (a-b)^2>0


which is true, so is that the correct proof and it holds true for all positive values of a and b?
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zdo0o
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imagine they weren't - show that comething crazy happens
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SimonM
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(Original post by RMorley)
Thanks, in the end I get to (a-b)^2>0


which is true, so is that the correct proof and it holds true for all positive values of a and b?
Not quite, you will need to show all steps are reversible, but yes, that's exactly what you want
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zdo0o
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(Original post by RMorley)
Thanks, in the end I get to (a-b)^2>0


which is true, so is that the correct proof and it holds true for all positive values of a and b?
as far as i'm aware, no. that only means that *if* it was true, then that is true - the fact that your second statement is true says nothign about the truth of the original statement.

to get around this, assume it was false, and find something clearly wrong (as opposed to assuming it was true, and finding something clearly true).

i might be wrong there, in which case i apologise, but i think that's the way to do it
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RMorley
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Oh I see, if I do it by contradiction I get (a-b)^2<0 and then if you do b^2-4ac it shows that, that statement cannot be true, therefore the arithmetic is greater then the geometric.

Thanks for your help
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