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I am stuck on this question:

given $a_1a_2≤(\frac{a_1+a_2}{2})^2$ prove by induction of m that $$a_1a_2...a_p≤(\frac{a_1+a_2+...+a_p}{p})^p$$ where $a_i$ are all positive and real and $p=2^m$ (an increase in m unity doubles the number of factors in the product)

On looking on other questions I know the sort of method but can't get it to work with this. I am fine with normal proof by induction (i.e. equals rather then inequalities) . Please can you give me some hints to work it out.

1 Answers1

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Note that

$$(a_1a_2)(a_3a_4)\leq\left[\frac{a_1+a_2}{2}\right]^2\left[\frac{a_3+a_4}{2}\right]^2 = \left[\left(\frac{a_1+a_2}{2}\right)\left(\frac{a_3+a_4}{2}\right)\right]^2\\ \leq \left[\frac1{4}\sum_{i=1}^{4}a_i\right]^4.$$

Use this to prove by induction that

$$\left[\prod_{i=1}^{2^n}a_i\right]^{1/2^n} \leq \frac1{2^n}\sum_{i=1}^{2^n}a_i.$$

RRL
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  • Thanks for what you have given me, I can do it from here, but can you please explain where the second ≤ on the first line came from? –  Jun 28 '14 at 16:26
  • I added a step -- apply the two term inequality $XY \leq [(X+Y)/2]^2$ where $X = (a_1+a_2)/2$ and $Y = (a_3+a_4)/2$ – RRL Jun 28 '14 at 16:40