If we have that $\frac{1}{1+x_1}+\frac{1}{1+x_2}+...+\frac{1}{1+x_n}=1$, then prove that $\sqrt[n]{x_1x_2...x_n}\ge (n-1)$. Where $x_1,x_2,...,x_n$ are all non negative real numbers.
I got $x_1+x_2+...+x_n\ge n(n-1)$. Also we have $\frac{x_1}{1+x_1}+\frac{x_2}{1+x_2}+...+\frac{x_n}{1+x_n}=n-1$.
let $$\dfrac{1}{1+x_{i}}=b_{i}\Longrightarrow x_{i}=\dfrac{1-b_{i}}{b_{i}},b_{1}+b_{2}+\cdots+b_{n}=1$$ Use AM-GM inequality $$b_{2}+b_{3}+\cdots+b_{n}\ge (n-1)\cdot\sqrt[n-1]{b_{2}b_{3}\cdots b_{n}}$$ $$b_{3}+b_{4}+\cdots+b_{1}\ge (n-1)\cdot\sqrt[n-1]{b_{3}b_{4}\cdots b_{1}}$$ $$\cdots\cdots$$ $$b_{1}+b_{2}+\cdots+b_{n-1}\ge (n-1)\cdot\sqrt[n-1]{b_{1}b_{2}\cdots b_{n-1}}$$ so we have $$x_{1}x_{2}\cdots x_{n}=\dfrac{\displaystyle\prod_{cyc}(b_{2}+b_{3}+\cdots+b_{n})}{\prod_{cyc}b_{1}}\ge \dfrac{(n-1)^{n}\cdot\prod_{cyc}\sqrt[n-1]{b_{2}\cdots b_{n}}}{\prod_{cyc}b_{1}}=(n-1)^n$$ so $$\sqrt[n]{x_{1}x_{2}\cdots x_{n}}\ge n-1$$
Another solution: $$\frac{1}{1+x_1}+\frac{1}{1+x_2}+...+\frac{1}{1+x_n}=1$$ implies $$\frac{x_k}{1+x_k}=1-\frac{1}{1+x_k}=\sum_{i=1,i\neq{k}}^n\frac{1}{1+x_i}\geq\frac{n-1}{\sqrt[n-1]{\prod_{i=1,i\neq{k}}^n(1+x_i)}}.$$
Multiplying these inequalities member from member for $k=1...n$ gives
$$\frac{x_1.x_2...x_n}{(1+x_1)(1+x_2)...(1+x_n)}\geq\frac{(n-1)^n}{(1+x_1)(1+x_2)...(1+x_n)}$$ hence the required inequality.
Other solution:
The cases $n=1,2$ are trivial. In the following, I consider $n\geq 3$.
Lemma: Given positive numbers $y_1$, $y_2$,..., $y_n$ and $d$ with $y_1+\cdots +y_n=1$, we have $$\left(\frac{1}{y_1+d}-1\right)\left(\frac{1}{y_2+d}-1\right)\cdots\left(\frac{1}{y_n+d}-1\right)\geq\left(\frac{1}{\frac{1}{n}+d}-1\right)^n\tag{1}$$
Proof: Let $d>0$ and $U=\{y=(y_1,y_2,...,y_n)\in\mathbb{R}^n;\;y_1>0,\;y_2>0,\dots, \;y_n>0\}$.
Define $f,\varphi:U\to\mathbb{R}$ by $f(y)=\left(\frac{1}{y_1+d}-1\right)\cdots\left(\frac{1}{y_n+d}-1\right)$ and $\varphi(y)=y_1+\cdots+y_n$.
It follows from the Lagrange Multipliers Method that:
$$y\in \varphi^{-1}(1)\text{ is a critical point of } f\quad\Longleftrightarrow\quad\nabla f(y)=\lambda \nabla \varphi(y) \text{ for some } \lambda\in\mathbb{R}$$ $$\Longleftrightarrow \qquad\frac{1}{Y_i^2}\left(\frac{1}{Y_j}-1\right) =\frac{1}{Y_j^2}\left(\frac{1}{Y_i}-1\right),\quad\forall\ i,j=1,...,n$$
$$\Longleftrightarrow \qquad y_1=y_2=\cdots=y_n=\frac{1}{n}\quad \text{(here we use $n\geq 3$)}$$
So, $q=(\frac{1}{n},\frac{1}{n},\dots,\frac{1}{n})$ is the unique critical point of $f$ in $\varphi^{-1}(1)$. It is possible to prove that $f$ attains minimum at $q$ (the argument is similar to this one).
Therefore, for all $x\in\varphi^{-1}(1)$ we have $$f(y)\geq f\left(\tfrac{1}{n},\tfrac{1}{n},\dots,\tfrac{1}{n}\right)\tag{2}$$ As $(\text{2})$ is $(1)$, the proof is complete.
Corollary 1: Given positive numbers $y_1$, $y_2$,..., $y_n$ with $y_1+\cdots +y_n=1$, we have $$\left(\frac{1}{y_1}-1\right)\left(\frac{1}{y_2}-1\right)\cdots\left(\frac{1}{y_n}-1\right)\geq\left(n-1\right)^n\tag{3}$$
Proof: Letting $d\to 0$ in $(1)$ we get $(3)$.
Corollary 2: Let $x_1,x_2,...,x_n>0$ with $\displaystyle\frac{1}{1+x_1}+\frac{1}{1+x_2}\cdots +\frac{1}{1+x_n}=1$. Then $$\sqrt[n]{x_1x_2\cdots x_n}\geq n-1$$
Proof: Apply Corollary 1 with $$y_1=\frac{1}{1+x_1},\quad y_2=\frac{1}{1+x_2},\quad\dots\quad y_n=\frac{1}{1+x_n}$$
