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consider positive numbers $a_1,a_2,a_3,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$. does the following in-equality holds and if it does then how to prove it

$\left[(a_1+b_1)(a_2+b_2)\cdots(a_n+b_n)\right]^{1/n}\ge \left(a_1a_2\cdots a_n\right)^{1/n}+\left(b_1b_2\cdots b_n\right)^{1/n}$

  • 1
    FYI, this is known as [Mahler's Inequality](http://en.wikipedia.org/wiki/Mahler%27s_inequality)2012-06-24

3 Answers 3

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Apply the AM-GM inequality to the sequence $a_k/(a_k+b_k)$ and then to the sequence $b_k/(a_k+b_k)$. Add the resulting two inequalities, and multiply through by $\left(\prod_k (a_k+b_k)\right)^{1/n}$ to get the result.

This is exercise 2.11 (page 34) of J. Michael Steele's The Cauchy-Schwarz Master Class, and the result is there credited to Minkowski. This inequality is sometimes called the "superadditivity of the geometric mean".

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The inequality does not hold.

Set $n=1$: the inequality reads $a + b \geq a b.$ With $a=2$ and $b=4$ the inequality is violated.

Why did you think the inequality does hold?

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    It was edited an hour after Fabian answered2011-03-28
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By Holder $\prod_{i=1}^n(a_i+b_i)\geq\left(\sqrt[n]{\prod_{i=1}^na_i}+\sqrt[n]{\prod_{i=1}^nb_i}\right)^n$ and we are done!