The eleventh question: Given a positive integer $n \geq 2$, find the smallest real number $c(n)$ such that for any positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$, the following inequality holds: $\frac{\sum_{i=1}^{n} a_{i}}{n}-\sqrt[n]{\prod_{i=1}^{n} a_{i}} \leq c(n) \cdot \max _{1 \leq i<j \leq n}\left\{\left(\sqrt{a_{i}}-\sqrt{a_{j}}\right)^{2}\right\}$.
The 11th question:
Solution: On one hand, we take $a_{1}=a_{2}=\ldots=a_{n-1}=1, a_{n}=\varepsilon$. Let $\varepsilon \rightarrow 0^{+}$, then we get $\frac{n-1}{n} \leq c(n)$.
Next, we prove that for any positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$, we have
$$
\frac{\sum_{i=1}^{n} a_{i}}{n}-\sqrt[n]{\prod_{i=1}^{n} a_{i}} \leq
\frac{n-1}{n} \cdot \max _{1 \leq i<j \leq n}\left\{\left(\sqrt{a_{i}}-\sqrt{a_{j}}\right)^{2}\right\}.
$$

Without loss of generality, assume $a_{1}=\max_{1 \leq i \leq n} a_{i}, a_{n}=\min_{1 \leq i \leq n} a_{i}$, then the inequality is equivalent to:
$$
\frac{\sum_{i=1}^{n} a_{i}}{n}-\sqrt[n]{\prod_{i=1}^{n} a_{i}} \leq \frac{n-1}{n} \cdot\left(\sqrt{a_{n}}-\sqrt{a_{1}}\right)^{2}=\frac{n-1}{n} \cdot\left(a_{1}+a_{n}-2 \sqrt{a_{1} a_{n}}\right)
$$

Let $F\left(a_{1}, a_{2}, \ldots, a_{n}\right)=\frac{\sum_{i=1}^{n} a_{i}}{n}-\sqrt[n]{\prod_{i=1}^{n} a_{i}}$.
Consider the function
$$
f(x)=F\left(a_{1}, a_{2}, \ldots, a_{i-1}, x, a_{i+1}, \ldots, a_{n}\right)=\frac{x}{n}+\frac{\sum_{j \neq i} a_{j}}{n}-\sqrt[n]{\sum_{j \neq 1} a_{j} \cdot x^{\frac{1}{n}}}
$$
$$
\begin{array}{l}
F\left(a_{1}, a_{2}, \ldots, a_{n}\right) \\
\leq \max \left\{F\left(a_{1}, \ldots, a_{i-1}, a_{1}, a_{i+1}, \ldots, a_{n}\right), F\left(a_{1}, \ldots, a_{i-1}, a_{n}, a_{i+1}, \ldots, a_{n}\right)\right\}
\end{array}
$$

Thus, we have
$$
\begin{array}{l}
F\left(a_{1}, a_{2}, \ldots, a_{n}\right) \\
\leq \max _{t \in Z, 1 \leq t \leq n-1}\{F(\underbrace{a_{1}, a_{1}, \ldots, a_{1}}_{t \uparrow}, \underbrace{a_{n}, a_{n}, \ldots, a_{n}}_{n-t \uparrow})\} \\
=\max _{t \in Z, 1 \leq t \leq n-1}\left\{\frac{t a_{1}+(n-t) a_{n}}{n}-a_{1}^{\frac{t}{n}} \cdot a_{n}^{\frac{n-t}{n}}\right\}
\end{array}
$$

Therefore, we only need to prove:
$$
\begin{array}{l}
\frac{t a_{1}+(n-t) a_{n}}{n}-a_{1}^{\frac{t}{n}} \cdot a_{n}^{\frac{n-t}{n}} \leq \frac{n-1}{n}\left(a_{1}+a_{n}-2 \sqrt{a_{1} a_{n}}\right) \\
\Leftrightarrow(n-t-1) a_{1}+(t-1) a_{n}+n \cdot a_{1}^{\frac{t}{n}} \cdot a_{n}^{\frac{n-t}{n}} \geq 2(n-1) \sqrt{a_{1} a_{n}}
\end{array}
$$

In fact, according to the mean inequality, we know:
$$
\begin{array}{l}
(n-t-1) a_{1}+(t-1) a_{n}+n \cdot a_{1}^{\frac{t}{n}} \cdot a_{n}^{\frac{n-t}{n}} \\
\geq(2 n-2) \cdot\left[a_{1}^{n-t-1} \cdot a_{n}^{t-1} \cdot\left(a_{1}^{\frac{t}{n}} \cdot a_{n}^{\frac{n-t}{n}}\right)^{n}\right]^{\frac{1}{2 n-2}}=2 n \sqrt{a_{1} a_{n}} \\
\end{array}
$$

Hence, (3) holds.
In summary, the required $c(n)=\frac{n-1}{n}$.