# Product of averages

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.

• $\frac{x_k+y_k}{2} \ge \sqrt{x_k y_k}$, so ... – Martin R Sep 28 at 7:55

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.

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

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