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 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 October 6th, 2019, 09:26 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 Fast derivation of am-gm let $\displaystyle s_n = \sum x_i /n$ and $\displaystyle g_n =\sqrt[n]{\prod x_i }$. set $\displaystyle f(x)=s_n /g_n =s_n / (e^{1/n \ln(s_n )} )$. $\displaystyle f(x) \equiv y/\sqrt[n]{e^{ln(y)}}\geq y/e^{ln(y)}\geq 1$. $\displaystyle f(x)>1 \Rightarrow s_n \geq g_n$. October 6th, 2019, 10:27 AM #2 Senior Member   Joined: Jun 2019 From: USA Posts: 310 Thanks: 162 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. October 6th, 2019, 11:12 AM #3 Senior Member   Joined: Jun 2019 From: USA Posts: 310 Thanks: 162 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. October 6th, 2019, 02:53 PM #4 Senior Member   Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 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. Thanks from idontknow October 6th, 2019, 11:32 PM   #5
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Quote:
 Originally Posted by tahirimanov19 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. October 7th, 2019, 06:28 AM   #6
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Math Focus: Area of Circle
Quote:
 Originally Posted by idontknow 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. Tags amgm, derivation, fast Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Pumaftw Elementary Math 2 October 27th, 2015 07:09 AM ahmed hussein New Users 0 April 27th, 2014 10:52 PM Solarmew Applied Math 8 April 27th, 2012 01:08 PM eric3353 Advanced Statistics 0 August 19th, 2008 03:46 PM

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