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Sorry if it is sort of hard to read so here it is in words.
Show that the nth root of the product of n terms is less than or equal to the sum of n terms divided by n.

Our instructions are to use a weird form of induction. Assuming it's true for k we are supposed to first show that it is then true for 2k then show it's true for k-1 to "fill in the gaps". I have found out how to show it's true for k-1 but I'm stuck trying to show it's true for 2k. Any advise would be appreciated. Thanks

jameselmore
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grizzle
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1 Answers1

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Use the induction on $n$. For $n=1,2$, the AM-GM inequality holds immediately. Now, suppose that for $n=2^r$, where $r\in\mathbb{N}$, the AM-GM inequality holds. That is, $$\frac{a_1+a_2+\cdots+a_{2^r}}{2^r}\geq\sqrt[2^r]{a_1a_2\cdots a_{2^r}}.$$ Then consider $n=2^{r+1}$, we have \begin{align*} \frac{a_1+a_2+\cdots+a_{2^{r+1}}}{2^{r+1}} &=\frac{1}{2}\left(\frac{a_1+\cdots+a_{2^r}}{2^r} +\frac{a_{2^{r}+1}+\cdots+a_{2^{r+1}}}{2^r}\right)\\ &\geq\frac{1}{2}\left(\sqrt[2^r]{a_1\cdots a_{2^{r}}} +\sqrt[2^r]{a_{2^{r}+1}\cdots a_{2^{r+1}}}\right)\\ &\geq\sqrt{\sqrt[2^r]{a_1\cdots a_{2^{r}}} \,\,\sqrt[2^r]{a_{2^{r}+1}\cdots a_{2^{r+1}}}}\\ &=\sqrt[2^{r+1}]{a_1a_2\cdots a_{2^{r+1}}}. \end{align*} So we conclude that the AM-GM inequality holds for all $n$ which is a power of $2$. Next, given $n\in\mathbb{N}$ such that $2^{r-1}<n<2^{r}$ for some $r\in\mathbb{N}$, we let $2^r=n+s$, where $0<s<2^{r-1}$. Then given positive reals $a_1,a_2,\ldots,a_n$, and let $k=\frac{a_1+a_2+\cdots+a_n}{n}$, we see that by the preceding result, \begin{align*} & &\frac{a_1+a_2+\cdots+a_n+sk}{n+s}&\geq\sqrt[n+s]{a_1a_2\cdots a_nk^s}\\ \Longleftrightarrow& &\left(\frac{nk+sk}{n+s}\right)^{n+s} &\geq a_1a_2\cdots a_nk^s\\ \Longleftrightarrow& & k^{n+s}&\geq a_1a_2\cdots a_nk^s\\ \Longleftrightarrow& & k^n&\geq a_1a_2\cdots a_n\\ \Longleftrightarrow& &\left(\frac{a_1+a_2+\cdots+a_n}{n}\right)^n&\geq a_1a_2\cdots a_n\\ \Longleftrightarrow& &\frac{a_1+a_2+\cdots+a_n}{n}&\geq \sqrt[n]{a_1a_2\cdots a_n}.\\ \end{align*} Thus we conclude that the AM-GM inequality holds for all $n\in\mathbb{N}$.

Solumilkyu
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