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Question: Let $x_i\in(0,\frac12]$ for $1\leq i\leq n$. Show that $$\sqrt[n]{x_1x_2\dots x_n}+\sqrt[n]{(1-x_1)(1-x_2)\dots(1-x_n)}\leq 1$$ without using optmization techniques in calculus.

My try: For $n=1$, it is trivially true. For $n=2$, let $x_1=u^2,x_2=v^2$. Then, $$\sqrt{1-u^2-v^2+u^2v^2}\leq 1-uv$$ then after squaring $$0\leq (u-v)^2.$$ At $n=3$, I am stuck. Is there an easier method to prove this inequality? Also, what is the lower bound of this sum? Thanks in advance.

Michael Rozenberg
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Bob Dobbs
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  • See also https://en.wikipedia.org/wiki/Mahler%27s_inequality – Martin R Jul 09 '23 at 13:43
  • What is the lower bound? There was a second part of question. – Bob Dobbs Jul 09 '23 at 13:44
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    The upper bound is what you are asking for in the title, and what has been asked and answered before (in fact more than once: https://math.stackexchange.com/questions/linked/29357). I would suggest that you ask a separate question for the lower bound. – Martin R Jul 09 '23 at 13:58
  • I think that Mahler's inequality also helps to determine the miminal value and I might have a solution for that. But again, it is better to ask that separately IMHO. – Martin R Jul 09 '23 at 15:23

2 Answers2

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By using AM-GM inequality, $$\sqrt[n]{x_1x_2\dots x_n}+\sqrt[n]{(1-x_1)(1-x_2)\dots(1-x_n)}\leq \frac{x_1+\ldots+x_n}{n} + \frac{n-(x_1+\ldots+x_n)}{n} = 1$$

Torrente
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Because by Holder: $$1=\sqrt[n]{\prod_{i=1}^n(x_i+1-x_i)}\geq\sqrt[n]{\prod_{i=1}^nx_i}+\sqrt[n]{\prod_{i=1}^n(1-x_i)}.$$

Michael Rozenberg
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