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the exercise says: If $a_1,\ldots,a_n\ge0$ then the A-M is $A_n=\frac{a_1+\cdots+a_n}{n}$ and G-M is $G_n=\sqrt[n]{a_1a_2\cdots a_n}$ we would like to show that $G_n\le A_n$

(1) suppose that $a_1<A_n$ then some $a_i>A_n$, for convenience say $a_2>A_n$ Let $\overline{a}_1=A_n$ and $\overline{a}_2=a_1+a_2-\overline{a}_1$ show that $$\overline{a}_1\overline{a}_2\ge a_1a_2$$

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I can't understand how he concludes $G_n\le \overline{G}_n$, maybe he says suppose that $a_3<A_n$ then for convenience say $a_4>A_n$, and so $$\overline{a}_1\overline{a}_2\ge a_1a_2\\ \overline{a}_3\overline{a}_4\ge a_3a_4\\ \vdots \\ \overline{a}_{n-1}\overline{a}_n\ge a_{n-1}a_n$$ and multiply each to get the result but what if $a_3=A_n$? is there a more rigorous proof? also i can't understand the rest of what he says if anyone can clarify it

thanks in advance

user153330
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  • i already saw this related question https://math.stackexchange.com/questions/384791/arithmetic-mean-is-less-than-geometric-mean-spivak-calculus-3rd-chapter-2-probl not the same as mine – user153330 Apr 07 '15 at 15:57
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    Set $\overline{a}_k = a_k$ for $3 \leqslant k \leqslant n$. Then $$\overline{a}_1\overline{a}_2 > a_1 a_2 \implies \overline{a}_1\overline{a}_2\overline{a}_3\dotsc \overline{a}_n > a_1 a_2 a_3 \dotsc a_n,$$ and therefore $$\overline{G}_n = \sqrt[n]{\overline{a}_1\overline{a}_2\overline{a}_3\dotsc \overline{a}_n} > \sqrt[n]{a_1 a_2 a_3 \dotsc a_n} = G_n.$$ – Daniel Fischer Sep 16 '15 at 21:56

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