Saudi Arabian Mathematical Competitions

Let $y_{k}=\frac{1}{1+x_{k}},\ k=1,2, \ldots, n$. Then
$$x_{k}=\frac{1}{y_{k}}-1=\frac{y_{1}+\ldots+y_{n}}{y_{k}}-1=\frac{S_{k}}{y_{k}}$$
where
$$S_{k}=x_{1}+\ldots+x_{k-1}+x_{k+1}+\ldots+x_{n}, \quad k=1,2, \ldots, n.$$
Applying AM-GM inequality it follows
$$\frac{S_{k}}{n-1} \geq \sqrt[n-1]{y_{1} \ldots y_{k-1} y_{k+1} \ldots y_{n}}$$
for $k=1,2, \ldots, n$. We obtain
$$x_{1} x_{2} \ldots x_{n}=\frac{S_{1} S_{2} \ldots S_{n}}{y_{1} y_{2} \ldots y_{n}} \geq (n-1)^{n} \frac{y_{1} y_{2} \ldots y_{n}}{y_{1} y_{2} \ldots y_{n}}=(n-1)^{n}.$$
The equality holds if and only if $y_{1}=\ldots=y_{n}=\frac{1}{n}$ that is $x_{1}=\ldots=x_{n}=n-1$.

Write the condition in the hypothesis as
$$\frac{1}{1+x_{1}}+\ldots+\frac{1}{1+x_{n-1}}=\frac{x_{n}}{1+x_{n}}$$
Applying AM-GM inequality we get
$$(n-1) \sqrt[n-1]{\frac{1}{1+x_{1}} \cdots \frac{1}{1+x_{n-1}}} \leq \frac{x_{n}}{1+x_{n}} \tag{1}$$
In similar way we obtain other $n-1$ inequalities of the form. Therefore, we have
$$(n-1) \sqrt[n-1]{\frac{1}{1+x_{1}} \cdots \frac{1}{1+x_{k-1}} \frac{1}{1+x_{k+1}} \cdots \frac{1}{1+x_{n}}} \leq \frac{x_{k}}{1+x_{k}} \tag{2}$$
for $k=1,2, \ldots, n$. Multiplying inequalities (2) and simplifying by $\frac{1}{1+x_{1}} \ldots \frac{1}{1+x_{n}}$ it follows $(n-1)^{n} \leq x_{1} \ldots x_{n}$.