SOLUTION. The inequality can be rewritten in the form
$$\sum_{i=1}^{n} a_{i}+n(n-1) \sqrt[n]{\prod_{i=1}^{n} a_{i}} \geq(n-1) \sum_{i=1}^{n} \sqrt[n-1]{\prod_{j \neq i} a_{j}}$$

First we will prove the following result (that helps us prove the previous inequality immediately): if $x_{1}, x_{2}, \ldots, x_{n}$ are real numbers then $\left(\alpha^{*}\right) \gg\left(\beta^{*}\right)$ with
$$\begin{array}{c} (\alpha)=\left(x_{1}, x_{2}, \ldots, x_{n}, x, x, \ldots, x\right) \\ (\beta)=\left(y_{1}, y_{1}, \ldots, y_{1}, y_{2}, y_{2}, \ldots, y_{2}, \ldots, y_{n}, y_{n}, \ldots, y_{n}\right) \end{array}$$
where $x=\frac{1}{n}\left(x_{1}+x_{2}+\ldots+x_{n}\right),(\alpha)$ includes $n(n-2)$ numbers $x,(\beta)$ includes $n-1$ numbers $y_{k}(\forall k \in\{1,2, \ldots, n\})$, and each number $b_{k}$ is determined from $b_{k}=$ $\frac{n x-x_{i}}{n-1}$
Indeed, by the symmetric majorization criterion, we only need to prove that
$$\left|x_{1}\right|+\left|x_{2}\right|+\ldots+\left|x_{n}\right|+(n-2)|S| \geq\left|S-x_{1}\right|+\left|S-x_{2}\right|+\ldots+\left|S-x_{n}\right| \text { (*) }$$
where $S=x_{1}+x_{2}+\ldots+x_{n}=n x$. In the case $n=3$, this becomes a well-known result
$$|x|+|y|+|z|+|x+y+z| \geq|x+y|+|y+z|+|z+x|$$

In the general case, assume that $x_{1} \geq x_{2} \geq \ldots \geq x_{n}$. If $x_{i} \geq S \forall i \in\{1,2, \ldots, n\}$ then
$$\text { RHS }=\sum_{i=1}^{n}\left(x_{i}-S\right)=-(n-1) S \leq(n-1)|S| \leq \sum_{i=1}^{n}\left|x_{i}\right|+(n-2)|S|=\text { LHS }$$
and the conclusion follows. The case $x_{i} \leq S \forall i \in\{1,2, \ldots, n\}$ is proved similarly. We consider the final case. There exists an integer $k(1 \leq k \leq n-1)$ such that $x_{k} \geq S \geq$ $x_{k+1}$. In this case, we can prove $(\star)$ simply as follows
$$\begin{aligned} \mathrm{RHS}= & \sum_{i=1}^{k}\left(x_{i}-S\right)+\sum_{i=k+1}^{n}\left(S-x_{i}\right)=\sum_{i=1}^{k} x_{i}-\sum_{i=k+1}^{n} x_{k+1}+(n-2 k) S, \\ & \leq \sum_{i=1}^{n}\left|x_{i}\right|+(n-2 k)|S| \leq \sum_{i=1}^{n}\left|x_{i}\right|+(n-2)|S|=\text { LHS } \end{aligned}$$
which is also the desired result. The problem is completely solved.