Saudi Arabia Mathematical Competitions

Let $y_k=\frac{1}{1+x_k},k=1,2,\dots ,n$. Then $$x_k=\frac{1}{y_k}-1=\frac{y_1+\dots +y_n}{y_k}-1=\frac{S_k}{y_k}$$ where $$S_k=x_1+\dots +x_{k-1}+x_{k+1}+\dots +x_n,k=1,2,\dots ,n.$$ Applying AM-GM inequality it follows $$\frac{S_k}{n-1}\ge \sqrt[n-1]{y_1\dots y_{k-1}{y}_{k+1}\dots y_n}$$ for $k=1,2,\dots ,n$. We obtain $$x_1{x}_2\dots x_n=\frac{S_1{S}_2\dots S_n}{y_1{y}_2\dots y_n}\ge (n-1{)}^n\frac{y_1{y}_2\dots y_n}{y_1{y}_2\dots y_n}=(n-1{)}^n.$$ The equality holds if and only if $y_1=\dots =y_n=\frac{1}{n}$ that is $x_1=\dots =x_n=n-1$.

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Alternative solution.

Write the condition in the hypothesis as $$\frac{1}{1+x_1}+\dots +\frac{1}{1+x_{n-1}}=\frac{x_n}{1+x_n}$$ Applying AM-GM inequality we get $$(n-1)\sqrt[n-1]{\frac{1}{1+x_1}\cdots \frac{1}{1+x_{n-1}}}\le \frac{x_n}{1+x_n}\text{\hspace{1em}(1)}$$ In similar way we obtain other $n-1$ inequalities of the form. Therefore, we have $$(n-1)\sqrt[n-1]{\frac{1}{1+x_1}\cdots \frac{1}{1+x_{k-1}}\frac{1}{1+x_{k+1}}\cdots \frac{1}{1+x_n}}\le \frac{x_k}{1+x_k}\text{\hspace{1em}(2)}$$ for $k=1,2,\dots ,n$. Multiplying inequalities (2) and simplifying by $\frac{1}{1+x_1}\dots \frac{1}{1+x_n}$ it follows $(n-1{)}^n\le x_1\dots x_n$.