How much larger is the AM than the GM?

To find $\mathbf{E}{\left(\frac{AM}{GM}\right)}$ , we will need to use integrals.

$\large\begin{aligned} \mathbf{E}{\left(\frac{AM}{GM}\right)}&=\frac{1}{n{\left(a_1-a_0\right)}^n}\overbrace{\int_{a_0}^{a_1}\int_{a_0}^{a_1}\cdots\int_{a_0}^{a_1}}^{n}\frac{x_1+x_2+\cdots+x_n}{\sqrt[n]{x_1x_2\cdots x_n}}dx_1dx_2\cdots dx_n\\ &=\frac{1}{n{\left(a_1-a_0\right)}^n}\overbrace{\int_{a_0}^{a_1}\int_{a_0}^{a_1}\cdots\int_{a_0}^{a_1}}^{n}\left(\frac{x_1}{\sqrt[n]{x_1x_2\cdots x_n}}+\frac{x_2}{\sqrt[n]{x_1x_2\cdots x_n}}+\cdots+\frac{x_n}{\sqrt[n]{x_1x_2\cdots x_n}}\right)dx_1dx_2\cdots dx_n\\ &=\frac{1}{n{\left(a_1-a_0\right)}^n}\overbrace{\int_{a_0}^{a_1}\int_{a_0}^{a_1}\cdots\int_{a_0}^{a_1}}^{n}\left(\frac{x_1^{1-\frac{1}{n}}}{\sqrt[n]{x_2x_3\cdots x_n}}+\frac{x_2^{1-\frac{1}{n}}}{\sqrt[n]{x_1x_3\cdots x_n}}+\cdots+\frac{x_n^{1-\frac{1}{n}}}{\sqrt[n]{x_1x_2\cdots x_{n-1}}}\right)dx_1dx_2\cdots dx_n\\ &=\frac{1}{n{\left(a_1-a_0\right)}^n}\overbrace{\int_{a_0}^{a_1}\int_{a_0}^{a_1}\cdots\int_{a_0}^{a_1}}^{n-1}\left [\frac{n}{2n-1}\frac{x_1^{2-\frac{1}{n}}}{\sqrt[n]{x_2x_3\cdots x_n}}+\frac{n}{n-1}\frac{x_1^{1-\frac{1}{n}}x_2^{1-\frac{1}{n}}}{\sqrt[n]{x_3x_4\cdots x_n}}+\cdots+\frac{n}{n-1}\frac{x_1^{1-\frac{1}{n}}x_n^{1-\frac{1}{n}}}{\sqrt[n]{x_2x_3\cdots x_{n-1}}}\right ]_{x_1=a_0}^{x_1=a_1}dx_2dx_3\cdots dx_n\\ &=\frac{1}{n{\left(a_1-a_0\right)}^n}\overbrace{\int_{a_0}^{a_1}\int_{a_0}^{a_1}\cdots\int_{a_0}^{a_1}}^{n-2}\left [\left [\frac{n}{2n-1}\frac{n}{n-1}\frac{x_1^{2-\frac{1}{n}}x_2^{1-\frac{1}{n}}}{\sqrt[n]{x_3x_4\cdots x_n}}+\frac{n}{2n-1}\frac{n}{n-1}\frac{x_1^{1-\frac{1}{n}}x_2^{2-\frac{1}{n}}}{\sqrt[n]{x_3x_4\cdots x_n}}+\cdots+\frac{n}{n-1}\frac{n}{n-1}\frac{x_1^{1-\frac{1}{n}}x_2^{1-\frac{1}{n}}x_n^{1-\frac{1}{n}}}{\sqrt[n]{x_2x_3\cdots x_{n-1}}}\right ]_{x_1=a_0}^{x_1=a_1}\right ]_{x_2=a_0}^{x_2=a_1}dx_3dx_4\cdots dx_n \end{aligned}$

(By the way, setting $a_0=0$ and then finding the limit of $a_1\to\infty$ will make the work infinitely easier, but we won't be able to find the general formula.)

It should become clear that a pattern is emerging. In fact, the above becomes $\large\begin{aligned} &=\frac{n^{n-1}}{{\left(a_1-a_0\right)}^n{\left(n-1\right)}^{n-1}\left(2n-1\right)}\left [ \left [ \cdots \left [\sum_{i=1}^{n}\left(\left(x_i^{2-\frac{1}{n}}\right)\prod_{j=1,j≠i}^{n}\left(x_j^{1-\frac{1}{n}}\right)\right)\right ]_{x_1=a_0}^{x_1=a_1} \right ]_{x_2=a_0}^{x_2=a_1} \cdots \right ]_{x_n=a_0}^{x_n=a_1}\\ &=\frac{n^{n-1}}{{\left(a_1-a_0\right)}^n{\left(n-1\right)}^{n-1}\left(2n-1\right)}n\left(a_1^{2-\frac{1}{n}}-a_0^{2-\frac{1}{n}}\right){\left(a_1^{1-\frac{1}{n}}-a_0^{1-\frac{1}{n}}\right)}^{n-1}\\ &=\frac{n^n\left(a_1^{2-\frac{1}{n}}-a_0^{2-\frac{1}{n}}\right){\left(a_1^{1-\frac{1}{n}}-a_0^{1-\frac{1}{n}}\right)}^{n-1}}{{\left(a_1-a_0\right)}^n{\left(n-1\right)}^{n-1}\left(2n-1\right)} \end{aligned}$

The last line is the formula for $\mathbf{E}{\left(\frac{AM}{GM}\right)}$ . Playing around with it shows that if one sets $a_0$ and $a_1$ closer, the value becomes closer to $1$ , as should be expected. Probably more interestingly is the fact that, if we set $n\to\infty$ , the value becomes $\frac{1}{2}\exp{\left(1-\frac{a_1\ln{a_1}-a_0\ln{a_0}}{a_1-a_0}\right)}\left(a_0+a_1\right)$ .

Now, back to $\mathbf{E}{\left(\frac{AM}{GM}\right)}=\frac{n^n\left(a_1^{2-\frac{1}{n}}-a_0^{2-\frac{1}{n}}\right){\left(a_1^{1-\frac{1}{n}}-a_0^{1-\frac{1}{n}}\right)}^{n-1}}{{\left(a_1-a_0\right)}^n{\left(n-1\right)}^{n-1}\left(2n-1\right)}$

Find the limit when $a_0\to 0, a_1\to \infty$ (setting $a_0=0$ and all factors with $a_1$ subsequently cancels out), and one obtains

$\frac{n^{n}}{{\left(n-1\right)}^{n-1}(2n-1)}$

Thus, $A=2,B=C=1$ . Therefore, $A+B+C=4$ .

This value grows increasingly closer to $\frac{1}{2}e$ for large $n$ . Thus, given a large list of positive real numbers that seems to be generated randomly from $[0,m]$ for some $m$ , one should expect that the arithmetic mean of the numbers is around $136\%$ that of the geometric mean. Else, for a large list of positive real numbers generated randomly from $[a_0,a_1]$ , one should expect this value to change to $\frac{1}{2}\exp{\left(1-\frac{a_1\ln{a_1}-a_0\ln{a_0}}{a_1-a_0}\right)}\left(a_0+a_1\right)$ .

Let $X_i\sim U(a_0,a_1)$ be our random variables for $i=1,2,...,n$ . Notice that they are independent and identically distributed as per our hypothesis. Then $E[\frac{AM}{GM}]=E\Big[\frac{X_1+X_2+...+X_n}{n\sqrt[n]{X_1X_2...X_n}}\Big]$ $\stackrel{(1)}{=}\frac{1}{n}\big[E\Big[\frac{X_1}{\sqrt[n]{X_1X_2...X_n}}\Big]+...+E\Big[\frac{X_n}{\sqrt[n]{X_1X_2...X_n}}\Big]\big]$ $\stackrel{(2)}{=}E\Big[\frac{X_1^{(n-1)/n}}{\sqrt[n]{X_2X_3...X_n}}\Big]$ $\stackrel{(3)}{=}E\big[X_1^{(n-1)/n}\big]\cdot E\big[\frac{1}{\sqrt[n]{X_2}}\big]\cdot ...\cdot E\big[\frac{1}{\sqrt[n]{X_n}}\big]$ $\stackrel{(4)}{=}E\big[X_1^{(n-1)/n}\big]\cdot E\big[\frac{1}{\sqrt[n]{X_1}}\big]^{n-1},$ where (1) uses linearity of the expected value operation, (2) uses the fact that $X_i$ are identically distributed, (3) uses independence, and (4) uses identical distribution.

Compute the first expected value: $E[X_1^{(n-1)/n}]=\int_{a_0}^{a_1}x^{(n-1)/n}\frac{1}{a_1-a_0}dx=\frac{n(a_1^{2-1/n}-a_0^{2-1/n})}{(a_1-a_0)(n-1)},$ and the second: $E[\frac{1}{\sqrt[n]{X_1}}]=\int_{a_0}^{a_1}\frac{1}{\sqrt[n]{x}}\frac{1}{a_1-a_0}dx=\frac{n(a_1^{1-1/n}-a_0^{1-1/n})}{(a_1-a_0)(n-1)},$ so that equation (4) above gives us $E[\frac{AM}{GM}]=\Big(\frac{n(a_1^{2-1/n}-a_0^{2-1/n})}{(a_1-a_0)(n-1)}\Big)\Big(\frac{n(a_1^{1-1/n}-a_0^{1-1/n})}{(a_1-a_0)(n-1)}\Big)^{n-1}.$ With this equality in mind, $\lim_{\substack{a_0\to 0 \\ a_1 \to \infty }}E[\frac{AM}{GM}]=\frac{n^n}{(2n-1)(n-1)^{n-1}},$ whence the desired value is $2+1+1=4$ .

EDIT: This suggests that, on average we should expect $AM/GM\approx\frac{n^n}{(2n-1)(n-1)^{n-1}}$ , where the means are taken of randomly selected positive real numbers. I thought it was pretty neat that, if we take this limit as $n\to\infty$ , we see $\frac{AM}{GM}\approx \frac{e}{2}$ if $n$ is large enough!

Having no idea how to do this Nick Turtle's way I decided to explore using Wolfram Alpha. There is no proof in this 'solution,' just following my nose to get an answer.

Using n=2 I did a double integral from 0 to 1 (both variables) of AM/GM and it gave $\frac{4}{3}$ .

Changing the integration limits to 0 to 2 gave 16/3, 0 to 3 gave 36/3. OK the upper limit doesn't matter since dividing by the area to give the expected value always gets us back to 4/3. No need to go to infinity! Red herring suspected.

So the hinted formula gives $\frac{2^{2}}{2A-B}=\frac{4}{3}$ or $2A-B=3$

Next n=3 and the triple integral from 0 to 1 (all 3 variables) gives $\frac{27}{20}$ . Hey, the numerator is $3^{3}$ . That's a good sign. (Integrating from 0 to 2 makes the answer 8 times bigger so Red herring almost certain.)

n=4 gives $\frac{256}{189}$ , n=5 gives $\frac{3125}{2304}$ The numerators work. Now to take a closer look at the denominators.

n=2, $2A-B=3$

n=3, $(3A-B)2^{3-C}=20=5*2^{2}$

n=4, $(4A-B)3^{4-C}=189=7*3^{3}$

n=5, $(5A-B)4^{5-C}=2304=9*4^{4}$

Which is more than enough for me to infer $A=2$ , $B=1$ , $C=1$ , and $A+B+C=1+1+2=\boxed{4}$ And it worked!

Nick's is way cooler (as much as I can follow.)

Oh my god!!!!

I just banged by keyboard and got this answer correct.🤔🤔