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if $a_1,a_2,a_3,\ldots,a_n≥0$ then the arithmetic mean is:

$A_n=\frac{a_1+a_2+a_3+\cdots+a_n}{n}$

and the geometric mean is:

$G_n=\sqrt[n]{a_1a_2a_3\ldots a_n}$

(a) Making use of the fact that $G_n≤A_n$ when $n=2$ prove by induction over $K$ that $G_n≤A_n$ for all $n=2^k$

(b)For a general n with $2^m<n$ .Aply the part (a) to the $2^m$ numbers.

$a_1,\ldots,a_n,\underbrace{A_n,\ldots,A_n}_{2^m-n\text{ times}}$

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  • $\begingroup$ What is your question? Are you stuck on (a) or (b)? What have you tried so far? $\endgroup$ – Sam Cappleman-Lynes Sep 18 '15 at 16:15
  • $\begingroup$ (b) I already prove when n=2 and for all n=2^k. $\endgroup$ – Iván Galeana Aguilar Sep 18 '15 at 16:21
  • $\begingroup$ What are you stuck on exactly? Can you show an attempt? $\endgroup$ – Sam Cappleman-Lynes Sep 18 '15 at 16:21
  • $\begingroup$ well I saw in Internet the prove for n=2k and the prove when n=(2k-1) I try to prove the general case when n=1, n=k and n=k+1, but it does not give me any idea about what should use for $2^m<n$ $\endgroup$ – Iván Galeana Aguilar Sep 18 '15 at 16:26
  • $\begingroup$ I think the question should say $n < 2^m$, not the other way round. $\endgroup$ – Sam Cappleman-Lynes Sep 18 '15 at 16:27

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