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October 6th, 2019, 09:26 AM   #1
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Fast derivation of am-gm

let $\displaystyle s_n = \sum x_i /n $ and $\displaystyle g_n =\sqrt[n]{\prod x_i
}$.
set $\displaystyle f(x)=s_n /g_n =s_n / (e^{1/n \ln(s_n )} )$.

$\displaystyle f(x) \equiv y/\sqrt[n]{e^{ln(y)}}\geq y/e^{ln(y)}\geq 1$.
$\displaystyle f(x)>1 \Rightarrow s_n \geq g_n $.
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October 6th, 2019, 10:27 AM   #2
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I mean, I've known since at least middle school that a square has more area than a rectangle with the same perimeter. That's enough to convince me that the arithmetic mean is going to be more than the geometric mean.

Now...
Was anyone going to tell me that $\displaystyle \sqrt[n]{\prod x_i}=e^{1/n \ln(\sum x_i/n )}$,
or did I have to learn that in a post about which mean is bigger?

Exactly how much did I miss out on by not learning maths on slide rules?

Last edited by DarnItJimImAnEngineer; October 6th, 2019 at 10:37 AM.
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October 6th, 2019, 11:12 AM   #3
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OK, your equation doesn't hold.
$\displaystyle ln \left( x_1 x_2 \right)^{1/2} = \frac{1}{2} \left( ln(x_1) + ln(x_2) \right) $
This doesn't generalise to a log of the sums.

If you need further proof the arithmetic and geometric means are independent:
$\displaystyle \vec{x} = (5,5), s = \frac{5+5}{2}=5, g=\sqrt{5\cdot5} = 5$
$\displaystyle \vec{x} = (3,7), s = \frac{3+7}{2}=5, g=\sqrt{3\cdot7} \approx 4.58$

Last edited by DarnItJimImAnEngineer; October 6th, 2019 at 11:14 AM.
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October 6th, 2019, 02:53 PM   #4
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Math Focus: Area of Circle
Is this topic about prooving $AM \ge GM$? If it is, there is a much simpler way of doing it.
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October 6th, 2019, 11:32 PM   #5
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Quote:
Originally Posted by tahirimanov19 View Post
Is this topic about prooving $AM \ge GM$? If it is, there is a much simpler way of doing it.
Yes. The hint is to apply $\displaystyle \frac{y}{y^{1/n}}\geq \frac{y}{y}\geq 1$ , where $\displaystyle y=s_n $ .

second method is starting from $\displaystyle (x+y)^{2} \geq 0 $ or $\displaystyle \frac{x+y}{2}\geq \sqrt{xy}$ , etc.

Last edited by idontknow; October 6th, 2019 at 11:40 PM.
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October 7th, 2019, 06:28 AM   #6
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Quote:
Originally Posted by idontknow View Post
Yes. The hint is to apply $\displaystyle \frac{y}{y^{1/n}}\geq \frac{y}{y}\geq 1$ , where $\displaystyle y=s_n $ .

second method is starting from $\displaystyle (x+y)^{2} \geq 0 $ or $\displaystyle \frac{x+y}{2}\geq \sqrt{xy}$ , etc.
OK. Then prove

$$\dfrac{a_1+a_2+...a_n}{n} \ge (a_1a_2...a_n)^{1/n}$$

using the second method.
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