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I have been reading this and this, but I don't understand how one of the steps works.

Let $a_n=\frac{a_1+a_2+\cdots+a_{n-1}}{n-1}.$

How do you set $a_n$ to meet certain criteria and not lose generality?

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user150246
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1 Answers1

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Let $P(n)$ state that the AM-GM inequality holds for the given $n$. The idea of this proof is to prove the following three statements:

  1. $P(1)$.
  2. $P(n) \longrightarrow P(2n)$.
  3. $P(n) \longrightarrow P(n-1)$ for $n\geq 1$.

Using the first two, you can prove by induction that $P(2^n)$ holds for all $n \geq 0$. The third then implies that $P(n)$ holds for all $n \geq 0$, since every integer is smaller than some power of two.

The step you are worried about is step 3. We are assuming the AM-GM inequality for $n$ and proving it for $n-1$. So given $a_1,\ldots,a_{n-1}$, we are free to choose any $b_1,\ldots,b_n$ we want and apply AM-GM. In this case, we choose $b_1 = a_1,\ldots,b_{n-1}=a_{n-1}$, and $b_n$ we set to the average of $a_1,\ldots,a_{n-1}$.

Yuval Filmus
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  • So we don't want $a_1, a_2, ..., a_{n-1}$ to be specific because we want them to work with any (n-1) sequences and not only in some special cases, but it doesn't matter with the nth element. Is that right? – user150246 Oct 09 '14 at 13:34
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    Right. We want to prove the AM-GM for $a_1,\ldots,a_{n-1}$, but when appealing to the AM-GM of length $n$, we can use whatever sequences we want. – Yuval Filmus Oct 09 '14 at 14:26