An Elementary Inequality Problem 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.

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.
