# If $\sum\limits_{i=1}^{2017}\frac{1}{x_i+2017}=\frac{1}{2017}$ for $x_i>0$, then prove that $\sqrt[2017]{x_1x_2…x_{2017}}\ge2016\cdot2017$ [closed]

Let $x_1,x_2,\cdots{},x_{2017}$ be positive reals such that $$\dfrac{1}{x_1+2017}+\cdots{}+\dfrac{1}{x_{2017}+2017}=\dfrac{1}{2017}.$$ Prove that: $$\sqrt[2017]{x_1x_2\cdots{}x_{2017}}\ge2016\cdot{}2017.$$

Progress: By the AM-HM inequality, I've managed to show that $$x_1+x_2+\cdots{}+x_{2017}\ge 2017^2(2016).$$ I'm not sure how to proceed further.

## closed as off-topic by Carl Mummert, Daniel W. Farlow, 2012ssohn, user91500, Claude LeiboviciNov 28 '16 at 7:17

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• Sounds like an interesting problem written verbatim. Do you have any progress or hints to tell us about? Are you stuck on this? Is this a challenge? – Kitter Catter Nov 27 '16 at 4:00
• Sounds like you can use that HM<=GM<=AM – Pat Devlin Nov 27 '16 at 4:23
• This is tagged "context-math" - please let us know the source of the problem. – Carl Mummert Nov 27 '16 at 23:08
• When does the contest end? That is, how soon do you want us to give you the answer? – Joel Reyes Noche Nov 28 '16 at 0:48

By AM-GM $\frac{1}{2017}-\frac{1}{x_i+2017}=\frac{x_i}{2017(x_i+2017)}=\sum\limits_{k\neq i}\frac{1}{x_k+2017}\geq\frac{2016}{\sqrt[2016]{\prod\limits_{k\neq i}(x_k+2017)}}$ and product of these
• $$\prod\limits_{i=1}^{2017} x_i \ge \prod\limits_{i=1}^{2017} \frac{2016\cdot 2017(x_i+2017)}{\sqrt[2016]{\prod\limits_{k\not =i} (x_k+2017)}}=(2016\cdot 2017)^{2017}$$ – Ricardo Largaespada Nov 27 '16 at 5:46