
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
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October 6th, 2019, 09:26 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98  Fast derivation of amgm
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.

October 6th, 2019, 11:32 PM  #5  
Senior Member Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98  Quote:
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  
Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle  Quote:
$$\dfrac{a_1+a_2+...a_n}{n} \ge (a_1a_2...a_n)^{1/n}$$ using the second method.  

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