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Given $n$ positive real numbers $x_1, x_2,...,x_{n+1}$, suppose: $$\frac{1}{1+x_1}+\frac{1}{1+x_2}+...+\frac{1}{1+x_{n+1}}=1$$ Prove:$$x_1x_2...x_{n+1}\ge{n^{n+1}}$$

I have tried the substitution $$t_i=\frac{1}{1+x_i}$$ The problem thus becomes:

Given $$t_1+t_2+...+t_{n+1}=1, t_i\gt0$$ Prove:$$(\frac{1}{t_1}-1)(\frac{1}{t_2}-1)...(\frac{1}{t_{n+1}}-1)\ge{n^{n+1}}$$ Which is equavalent to the following: $$(\frac{t_2+t_3+...+t_{n+1}}{t_1})(\frac{t_1+t_3+...+t_{n+1}}{t_2})...(\frac{t_1+t_2+...+t_{n}}{t_{n+1}})\ge{n^{n+1}}$$ From now on, I think basic equality (which says arithmetic mean is greater than geometric mean) can be applied but I cannot quite get there. Any hints? Thanks in advance.

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From last, let $\sqrt[n]{t_1t_2...t_{n+1}}=A$ with $AM-GM$: $$\frac{t_2+t_3+...+t_{n+1}}{n}\geq\sqrt[n]{t_2t_3...t_{n+1}}=\frac{A}{\sqrt[n]{t_1}}$$ so for the first $$t_2+t_3+...+t_{n+1}\geq\frac{nA}{\sqrt[n]{t_1}}$$ and for the rest take like this.

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  • $\begingroup$ Aha! Got it! Thank you very much! $\endgroup$ – ryan w. Jan 18 '17 at 16:47
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By AM-GM $$\prod\limits_{i=1}^{n+1}\frac{x_{i}}{x_i+1}=\prod\limits_{i=1}^{n+1}\sum\limits_{k\neq i}\frac{1}{x_k+1}\geq\prod\limits_{i=1}^{n+1}\frac{n}{\sqrt[n]{\prod\limits_{k\neq i}(1+x_k)}}=\frac{n^{n+1}}{\prod\limits_{i=1 }^{n+1}(1+x_i)}$$ and we are done!

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    $\begingroup$ Nicely done! Your proof is really skillful, thanks! $\endgroup$ – ryan w. Jan 18 '17 at 16:45
  • $\begingroup$ @ryan w. You are welcome! $\endgroup$ – Michael Rozenberg Jan 18 '17 at 16:45

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