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Let $(x_1,...,x_n)$ and $(y_1,...,y_n)$ be two different tuples of positive reals such that $x_1\times\dots\times x_n=y_1\times\dots\times y_n = c$. Is it true that $$\left(\frac{x_1+y_1}{2}\right)\times\cdots\times \left(\frac{x_n+y_n}{2}\right) > c?$$

I think this should follow from a concavity argument, perhaps on the function $f(x_1,...,x_n) = x_1\times\cdots\times x_n$, but not sure how exactly.

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    $\begingroup$ $\frac{x_k+y_k}{2} \ge \sqrt{x_k y_k}$, so ... $\endgroup$ – Martin R Sep 28 at 7:55
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The inequality between arithmetic and geometric mean states that $\frac{x+y}{2} \ge \sqrt{x y}$ for $x, y \ge 0$, with equality if and only if $x=y$.

It follows that $$ \prod_{k=1}^n \frac{x_k+y_k}{2} \ge \prod_{k=1}^n \sqrt{x_k y_k} = \sqrt{\prod_{k=1}^n x_k \cdot \prod_{k=1}^n y_k} = c $$ with equality if and only if $(x_1, \ldots, x_n) = (y_1, \ldots, y_n)$.

If you want to use a concavity argument then consider $$ g(x_1, \ldots, x_n) = \log x_1 + \ldots + \log x_n \, . $$ $g$ is concave as a sum of concave functions.

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By Holder $$\prod_{k=1}^n\frac{x_k+y_k}{2}\geq\frac{\left(\sqrt[n]{\prod\limits_{k=1}^nx_k}+\sqrt[n]{\prod\limits_{k=1}^ny_k}\right)^n}{2^n}=c$$

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  • $\begingroup$ Without the proof? $\endgroup$ – nilo de roock Sep 28 at 8:02
  • $\begingroup$ @nilo de roock It's the proof. What is your question? I am ready to explain. $\endgroup$ – Michael Rozenberg Sep 28 at 8:03
  • $\begingroup$ What do you mean by "By Holder"? $\endgroup$ – nilo de roock Sep 28 at 8:05
  • $\begingroup$ About Holder see here: math.stackexchange.com/edit-tag-wiki/5773 I used Holder for $n$ sequences: $(x_k,y_k)$ $\endgroup$ – Michael Rozenberg Sep 28 at 8:06
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    $\begingroup$ OK. Got it. Thank you. $\endgroup$ – nilo de roock Sep 28 at 8:07

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