2
$\begingroup$

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.

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Daniel W. Farlow, 2012ssohn, user91500, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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? $\endgroup$ – Kitter Catter Nov 27 '16 at 4:00
  • $\begingroup$ Sounds like you can use that HM<=GM<=AM $\endgroup$ – Pat Devlin Nov 27 '16 at 4:23
  • 1
    $\begingroup$ This is tagged "context-math" - please let us know the source of the problem. $\endgroup$ – Carl Mummert Nov 27 '16 at 23:08
  • $\begingroup$ When does the contest end? That is, how soon do you want us to give you the answer? $\endgroup$ – Joel Reyes Noche Nov 28 '16 at 0:48
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ $$\prod\limits_{i=1}^{2017} x_i \ge \prod\limits_{i=1}^{2017} \frac{2016\cdot 2017(x_i+2017)}{\sqrt[2016]{\prod\limits_{k\not =i} (x_k+2017)}}=(2016\cdot 2017)^{2017}$$ $\endgroup$ – Ricardo Largaespada Nov 27 '16 at 5:46

Not the answer you're looking for? Browse other questions tagged or ask your own question.