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We are supposed to prove that $ \frac{\sum_{k=1}^{n}x_k}{n}\geq (\prod_{k=1}^{n} x_k)^{1/n}$, when $x_i\geq 0$ for all $n\in\mathbb{N}$.

My friend proved it for $n=2$, $n=3$ and then for an arbitrary integer $n$. Now I only think the proof for an arbitrary integer $n\geq 2$ is enough. Here is the proof excluding the first two cases.

Let $c=x_1x_2... x_n$ where $x_i>0$ when $1\leq i \leq n$ for some $n\in\mathbb{N}$.Let $a_k=\frac{x_1x_2... x_k}{c^{k/n}}$ and $b_k=\frac{c^{k/n}}{x_1x_2... x_k}$ for $1\leq k \leq n$. Then we have that $a_kb_k=1$. If $1\leq k \leq n-1$ then $$a_{k+1}b_{k}=\left(\frac{x_1x_2... x_kx_{k+1}}{c^{(k+1)/n}}\right)\left(\frac{c^{k/n}}{x_1x_2... x_k}\right)=\frac{x_{k+1}}{c^{1/n}}. $$ Furhermore, $a_1b_n=\frac{x_1}{c^{1/n}}.$ Hence we have \begin{align*} &a_1b_1+\dotsb+a_n b_n \leq a_1b_n+a_2b_1+a_3b_2+\dotsb+a_n b_{n-1} (I.1)\\ &\Longleftrightarrow n \leq \frac{x_1}{c^{1/n}}+\frac{x_2}{c^{1/n}}+\dotsb+\frac{x_n}{c^{1/n}}\\ &\Longleftrightarrow \sqrt[n]{x_1\cdot\dotsc\cdot x_n}\leq \frac{x_1+\dotsb+x_n}{n}. \end{align*}

So I think there should be another case that proves the AM-GM inequality for sequences of non negative numbers where atleast one number is $0$. Also I dont understand how the inequality I.1 is true.

per persson
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  • "arbitrary sequence of positive numbers where atleast one number is $0.$" What do you think positive means? Is zero a positive number? – Thomas Andrews Feb 08 '22 at 23:44
  • The big mystery is $I.1.$ Presumably, $I.1$ is a reference to some other result? – Thomas Andrews Feb 08 '22 at 23:46
  • I know Thomas, thanks for your input. So I.1 is where it breaks down I guess. – per persson Feb 08 '22 at 23:49
  • It seems to me that you are using $c$ as a constant. The value of $c$ in $a_{k+1}$ is different than the one in $b_k$. – Andrei Feb 08 '22 at 23:59
  • Usually $AM\ge GM$ is stated for positive numbers, but easily extended to non-negative numbers: if at least one of the numbers is $0$ and the rest positive, then the arithmetic mean is non-negative and the geometric mean is $0$. – Henry Feb 09 '22 at 00:04
  • $b_k$ is the inverse of $a_k$ and c should be the same if $1\leq k \leq n$. I can't see how c is different. – per persson Feb 09 '22 at 00:05
  • Sorry, my mistake. I misunderstood. I thought that $c$ is the product only to $k$, not to $n$. – Andrei Feb 09 '22 at 00:09

1 Answers1

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Inequality $I.1$ is an instance of the rearrangement inequality: Given two finite sequences $a_1,a_2,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$, of real numbers, consider the sum of products of the form $$\sum_{i=1}^n a_i b_{\sigma(i)},\tag1$$ where $b_{\sigma(1)},b_{\sigma(2)},\ldots,b_{\sigma(n)}$ is a permutation of $b_1, b_2,\ldots,b_n$. Then the sum (1) is minimized when the largest $a$ is paired with the smallest $b$, the second largest $a$ is paired with the second smallest $b$, and so on until the smallest $a$ is paired with the largest $b$.

In your context this condition is satisfied, since you have $a_ib_i=1$ for every $i$. So by the rearrangement inequality, the RHS $a_1b_n+a_2b_1+a_3b_2+\cdots+a_nb_{n-1}$ is guaranteed to be at least as big as the LHS $a_1b_1+\cdots+a_nb_n$.

grand_chat
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  • That's really good. With this, the rest of the proof is easy I guess. But just to be sure. It's is not necessary to prove it when n=2 and n=3. And also the easy case when atleast one number in the sequence is 0 should be considered. – per persson Feb 09 '22 at 00:20
  • Yes, this proof works for $n=2$ and $n=3$. When at least one number is zero, then the product is zero, so dividing by $c$ is illegal. But this case is indeed easy to prove. – grand_chat Feb 09 '22 at 00:25