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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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Yo thanks for the edit dawg –  fdkjghre May 9 '11 at 15:05
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I'm neither a "Yo" nor a "dawg". –  Arturo Magidin May 9 '11 at 15:10
    
LOL yo is a greeting and dawg is just a term of endearment mang. I'm thanking you for hookin a brotha up with an edit that made my math look supa dope! –  fdkjghre May 9 '11 at 15:13
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And I'm telling you that I don't find your terms of endearment particularly appealing. Kindly show your appreciation by refraining from such expressions when addressing me, or by not addressing me at all in the future. It's not funny, it's not cute, and I don't find it appreciative. –  Arturo Magidin May 9 '11 at 15:15
    
Okay, sorry about that. I will keep my slang out of this website. –  fdkjghre May 9 '11 at 15:17
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2 Answers

  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.

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