The twelfth question: Given a positive integer $n \geq 2$, find the smallest positive real number $c$ such that for any positive real numbers $x_{1}, x_{2}, \ldots, x_{n}$, the following inequality always holds: $\frac{1}{n}\left(\sum_{i=1}^{n} x_{i}\right)-\sqrt[n]{\prod_{i=1}^{n} x_{i}} \leq c \cdot \sum_{1 \leq i<j \leq n}\left(\sqrt{x_{i}}-\sqrt{x_{j}}\right)^{2}$.
The twelfth question:
Solution: On the one hand, take $x_{1} \rightarrow 0^{+}, x_{2}=x_{2}=\ldots=x_{n}=1$, then
$$
\frac{1}{n} \cdot(n-1) \leq c \cdot(n-1) \Rightarrow c \geq \frac{1}{n}
$$

On the other hand, we prove: For any positive real numbers $x_{1}, x_{2}, \ldots, x_{n}$, it always holds that
$$
\frac{1}{n}\left(\sum_{i=1}^{n} x_{i}\right)-\sqrt[n]{\prod_{i=1}^{n} x_{i}} \leq \frac{1}{n} \cdot \sum_{1 \leq i<j \leq n}\left(\sqrt{x_{i}}-\sqrt{x_{j}}\right)^{2}
$$

In fact, according to the Lagrange identity: $\sum_{1 \leq i<j \leq n}\left(\sqrt{x_{i}}-\sqrt{x_{i}}\right)^{2}=n \cdot \sum_{i=1}^{n} x_{i}-\left(\sum_{i=1}^{n} \sqrt{x_{i}}\right)^{2}$, so we only need to prove:
$$
(n-1)\left(\sum_{i=1}^{n} x_{i}\right)+n \sqrt[n]{\prod_{i=1}^{n} x_{i}} \geq\left(\sum_{i=1}^{n} \sqrt{x_{i}}\right)^{2}
$$

In fact, according to the Suranyi inequality:
$$
(n-1)\left(\sum_{i=1}^{n} x_{i}\right)+n \sqrt[n]{\prod_{i=1}^{n} x_{i}} \geq\left(\sum_{i=1}^{n} x_{i}^{\frac{1}{n}}\right)\left(\sum_{i=1}^{n} x_{i}^{\frac{n-1}{n}}\right) \geq\left(\sum_{i=1}^{n} \sqrt{x_{i}}\right)^{2}
$$

Thus, (2) holds.
In summary, the minimum value of $c$ is $\frac{1}{n}$.
Note: (Suranyi inequality) Given a positive integer $n$, and $n$ non-negative real numbers $a_{1}, a_{2}, \ldots, a_{n}$, prove: $(n-1) \left(\sum_{i=1}^{n} a_{i}^{n}\right)+n \prod_{i=1}^{n} a_{i} \geq\left(\sum_{i=1}^{n} a_{i}\right)\left(\sum_{i=1}^{n} a_{i}^{n-1}\right)$.
Proof: Induction on $n$.
When $n=1, 2$, the conclusion is obviously true.
Assume the conclusion holds for $n$. Now consider the case for $n+1$.
Without loss of generality, assume $a_{n+1}=\min _{1 \leq i \leq n+1} a_{i}$, and set $\sum_{i=1}^{n} a_{i}=n$. We need to prove:
$$
n\left(\sum_{i=1}^{n} a_{i}^{n+1}\right)+n a_{n+1}^{n+1}+n a_{n+1} \prod_{i=1}^{n} a_{i}+a_{n+1} \prod_{i=1}^{n} a_{i}-\left(n+a_{n+1}\right)\left(\sum_{i=1}^{n} a_{i}^{n}+a_{n+1}^{n}\right) \geq 0
$$

According to the induction hypothesis:
$$
(n-1)\left(\sum_{i=1}^{n} a_{i}^{n}\right)+n \prod_{i=1}^{n} a_{i} \geq\left(\sum_{i=1}^{n} a_{i}\right)\left(\sum_{i=1}^{n} a_{i}^{n-1}\right)=n \sum_{i=1}^{n} a_{i}^{n-1}
$$
.... . . . . ... ...
$$
\begin{array}{l}
\Rightarrow n a_{n+1} \prod_{i=1}^{n} a_{i} \geq n a_{n+1}\left(\sum_{i=1}^{n} a_{i}^{n-1}\right)-(n-1) a_{n+1}\left(\sum_{i=1}^{n} a_{i}^{n}\right) \\
\quad \text { Thus, we only need to prove: } \\
\quad n\left(\sum_{i=1}^{n} a_{i}^{n+1}\right)+n a_{n+1}^{n+1}+n a_{n+1}\left(\sum_{i=1}^{n} a_{i}^{n-1}\right)-(n-1) a_{n+1}\left(\sum_{i=1}^{n} a_{i}^{n}\right)+a_{n+1} \prod_{i=1}^{n} a_{i}- \\
\left(n+a_{n+1}\right)\left(\sum_{i=1}^{n} a_{i}^{n}+a_{n+1}^{n}\right) \geq 0 \\
\quad \Leftrightarrow n\left(\sum_{i=1}^{n} a_{i}^{n+1}\right)+(n-1) a_{n+1}^{n+1}+n a_{n+1}\left(\sum_{i=1}^{n} a_{i}^{n-1}\right)-n a_{n+1}\left(\sum_{i=1}^{n} a_{i}^{n}\right)+ \\
a_{n+1} \prod_{i=1}^{n} a_{i}-n\left(\sum_{i=1}^{n} a_{i}^{n}\right)-n a_{n+1}^{n} \geq 0 \\
\quad \Leftrightarrow n\left(\sum_{i=1}^{n} a_{i}^{n+1}-\sum_{i=1}^{n} a_{i}^{n}\right)-n a_{n+1}\left(\sum_{i=1}^{n} a_{i}^{n}-\sum_{i=1}^{n} a_{i}^{n-1}\right)+a_{n+1}\left[(n-1) a_{n+1}^{n}+\right. \\
\left.\prod_{i=1}^{n} a_{i}-n a_{n+1}^{n-1}\right] \geq 0
\end{array}
$$

In fact, since $a_{n+1} \leq a_{i}(1 \leq i \leq n)$, and $\sum_{i=1}^{n} a_{i}=n$, hence
$$
\begin{array}{l}
(n-1) a_{n+1}^{n}+\prod_{i=1}^{n} a_{i}-n a_{n+1}^{n-1} \\
=(n-1) a_{n+1}^{n}+\prod_{i=1}^{n}\left(a_{n+1}+\left(a_{i}-a_{n+1}\right)\right)-n a_{n+1} n-1 \\
\geq(n-1) a_{n+1}{ }^{n}+\left[a_{n+1}^{n}+a_{n+1} n-1\left(\sum_{i=1}^{n}\left(a_{i}-a_{n+1}\right)\right)\right]-n a_{n+1} n-1 \\
=n a_{n+1} n+a_{n+1} n-1\left(n-n a_{n+1}\right)-n a_{n+1} n-1=0
\end{array}
$$

Also, according to the Chebyshev inequality: $\sum_{i=1}^{n} a_{i}^{n}-\sum_{i=1}^{n} a_{i}^{n-1}=\sum_{i=1}^{n} a_{i}^{n}-\frac{\left(\sum_{i=1}^{n} a_{i}\right)\left(\sum_{i=1}^{n} a_{i}^{n-1}\right)}{n} \geq 0$, and $a_{n+1} \leq \frac{\sum_{i=1}^{n} a_{i}}{n}=1$, hence
$$
\begin{array}{l}
n\left(\sum_{i=1}^{n} a_{i}^{n+1}-\sum_{i=1}^{n} a_{i}^{n}\right)-n a_{n+1}\left(\sum_{i=1}^{n} a_{i}^{n}-\sum_{i=1}^{n} a_{i}^{n-1}\right) \\
\geq n\left(\sum_{i=1}^{n} a_{i}^{n+1}-\sum_{i=1}^{n} a_{i}^{n}\right)-n\left(\sum_{i=1}^{n} a_{i}^{n}-\sum_{i=1}^{n} a_{i}^{n-1}\right) \\
=n \sum_{i=1}^{n} a_{i}^{n-1}\left(a_{i}-1\right)^{2} \geq 0
\end{array}
$$

By (2) and (3), (1) holds. The proposition is proved!