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I am trying to follow along on this proof of the Arithmetic-geometric mean inequality, but I pretty much crashed at a couple steps.

  • If $a_1 \leq G \leq a_n$, then why is it that $a_1 + a_n \geq \frac{a_1a_n}{G}+ G$?

  • Why do we remove $a_1$ in the induction hypothesis and put in $\frac{a_1a_n}{G}$ ?

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    Okay, sorry about that. I will keep my slang out of this website.2011-05-09

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  1. It is presented slightly strangely on the slide, but if $0 \lt a_1 \le G \le a_n$ then it should be simple to see that $\frac{1}{G}(G-a_1)(a_n-G) \ge 0$ and you can then multiply out and rearrange.
  2. The induction hypothesis allows you to use any $n-1$ positive terms, by its assumption
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The proof you linked to contains the answer to your first question. $a_1+a_n-G-\frac{a_1a_n}{G}=\frac{1}{G}(G-a_1)(a_n-G)\ge 0$ as all terms in the paranthesis are nonnegative (G is clearly non-zero since a_1>0). It follows that $a_1+a_n \ge G + \frac{a_1a_n}{G}$. For your second question, note that the induction hypothesis implies that the inequality in question holds for any collection of n-1 positive numbers, the n-1 numbers $a_{2},...a_{n-1},\frac{a_1a_n}{G}$ are used because it suits us to do so, in the proof.