Let $y_i = \frac{1}{1+x_i}$ for all $i$, then $y_1+y_2+...+y_{n+1} = 1$. Hence, $x_1x_2...x_{n+1} = \frac{(1-y_1)(1-y_2)...(1-y_{n+1})}{y_1y_2...y_{n+1}}$ $=\frac{(\sum_{i \neq 1} y_i)(\sum_{i \neq 2} y_i)...(\sum_{i \neq n+1} y_i)}{y_1y_2...y_{n+1}}$ $\geq \frac{[n (\prod_{i \neq 1} y_i)^\frac{1}{n}][n( \prod_{i \neq 2} y_i)^\frac{1}{n})]...[n (\prod_{i \neq n+1} y_i)^\frac{1}{n}]}{y_1y_2...y_{n+1}}$ $= \frac{n^{n+1} \prod y_i}{y_1y_2...y_{n+1}}$ $= n^{n+1}$, where the inequality is due to AM-GM, so there is equality when $y_1=y_2=...=y_{n+1}$, or equivalently, $x_1=x_2=...=x_{n+1}$.

xeroxia wrote: $x_1, x_2,\cdots,x_{n+1}$ are posive real numbers satisfying the equation $\frac{1}{1+x_1} + \frac{1}{1+x_2} + \cdots + \frac{1}{1+x_{n+1}} =1$ Prove that $x_1x_2 \cdots x_{n+1} \geq n^{n+1}$. Since \[\frac{x_1}{1+x_{1}}=\frac{1}{1+x_2}+\cdots+\frac{1}{1+x_{n+1}}\geq n\sqrt[n]{\prod_{i\neq 1}\frac{x_i}{1+x_i}};\] We get these analogous $n+1$ relations, which, on multiplying throughout, lead to \[\prod_{i=1}^n\frac{x_i}{1+x_i}\geq n^{n+1}\prod_{i=1}^n\sqrt[n]{\prod_{j\neq i}\frac{x_j}{1+x_j}};\] Or, \[x_1x_2\cdots x_n\geq n^{n+1}.\] We are done. $\Box$

xeroxia wrote: $x_1, x_2,\cdots,x_{n+1}$ are posive real numbers satisfying the equation $\frac{1}{1+x_1} + \frac{1}{1+x_2} + \cdots + \frac{1}{1+x_{n+1}} =1$ Prove that $x_1x_2 \cdots x_{n+1} \geq n^{n+1}$. Result follows immediately from the well known inequality: $1=\sum_{i=1}^{n+1}\frac{1}{1+x_i} \ge \frac{n+1}{1+\sqrt[n+1]{x_1\cdots x_{n+1}}}$ $\Leftrightarrow x_1\cdots x_{n+1} \ge n^{n+1}$ Proof of first inequality see here