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#### Reckoner

##### Member

- Jun 16, 2012

- 45

"Prove that

\[\frac1n\sum_{i=1}^nx_i \geq \left(\prod_{i=1}^nx_i\right)^{1/n}\]

for positive integers \(n\) and positive real numbers \(x_i\)."

The author notes, "It does not seem possible to give a direct proof of this result using induction on \(n\). However, it can be proved for \(n = 2^m\) for \(m\geq0\) by induction on \(m\). The general result now follows by proving the converse of the usual inductive step: if the result holds for \(n = k + 1\), where \(k\) is a positive integer, then it holds for \(n = k\)."

So, following the author's advice, I try to show that

\[\frac1{2^m}\sum_{i=1}^{2^m}x_i \geq \left(\prod_{i=1}^{2^m}x_i\right)^{1/2^m}\]

for nonnegative integers \(m\). The base case is straightforward. Here's what I've tried for the inductive step:

Assume

\[\frac1{2^k}\sum_{i=1}^{2^k}x_i \geq \left(\prod_{i=1}^{2^k}x_i\right)^{1/2^k}\]

for some \(k \geq 0\). Then

\[\frac1{2^{k+1}}\sum_{i=1}^{2^{k+1}}x_i = \frac12\left(\frac1{2^k}\sum_{i=1}^{2^k}x_i + \frac1{2^k}\sum_{i=2^k+1}^{2^{k+1}}x_i\right).\]

Letting \(j = i - 2^k,\) we have

\[\frac12\left(\frac1{2^k}\sum_{i=1}^{2^k}x_i + \frac1{2^k}\sum_{i=2^k+1}^{2^{k+1}}x_i\right) = \frac12\left(\frac1{2^k}\sum_{i=1}^{2^k}x_i + \frac1{2^k}\sum_{j=1}^{2^k}x_{j+2^k}\right)\]

\[\geq\frac12\left(\prod_{i=1}^{2^k}x_i\right)^{1/2^k} + \frac12\left(\prod_{j=1}^{2^k}x_{j+2^k}\right)^{1/2^k}\mbox{ (by inductive hypothesis)}\]

\[=\frac12\left(\prod_{i=1}^{2^k}x_i\right)^{1/2^k} + \frac12\left(\prod_{i=2^k+1}^{2^{k+1}}x_i\right)^{1/2^k}.\]

And I'm not sure where to go from here. All that would be left is to show that

\[\frac12\left(\prod_{i=1}^{2^k}x_i\right)^{1/2^k} + \frac12\left(\prod_{i=2^k+1}^{2^{k+1}}x_i\right)^{1/2^k}\geq \left(\prod_{i=1}^{2^{k+1}}x_i\right)^{1/2^{k+1}},\]

but I'm not seeing it. Maybe I need to employ a different strategy altogether. Any ideas?