For $c$ being a real number, find the minimum value of $c$ such that for any positive real numbers $a_{1}, a_{2}, a_{3}, a_{4}$, the following inequality holds:
$$
\begin{array}{l}
\frac{1}{4} \sum_{i=1}^{4} a_{i}-\sqrt[4]{a_{1} a_{2} a_{3} a_{4}} \\
\leqslant c \max _{1 \leqslant i<j \leqslant 4}\left\{\left(\sqrt{a_{i}}-\sqrt{a_{j}}\right)^{2}\right\} .
\end{array}
$$
Notice that for any $x_{1}, x_{2}, x_{3}, x_{4} > 0$, we have
$$
\begin{array}{l}
\sum_{i=1}^{4} x_{i}^{2}-4 \sqrt{x_{1} x_{2} x_{3} x_{4}} \\
=\left(x_{1}-x_{2}\right)^{2}+2\left(\sqrt{x_{1} x_{2}}-\sqrt{x_{3} x_{4}}\right)^{2}+\left(x_{3}-x_{4}\right)^{2} .
\end{array}
$$

Without loss of generality, assume $x_{1} \geqslant x_{2} \geqslant x_{3} \geqslant x_{4}$.
Then $\left(x_{1}-x_{4}\right)^{2}=\max _{1 \leqslant i0$, there exists $c<\frac{3}{4}$ such that the conclusion holds. Then take
$$
a_{1}=a_{2}=a_{3}=1, a_{4}=\frac{1}{n^{2}}(n \geqslant 1),
$$

Thus, the conclusion can be rewritten as
$$
\frac{1}{4}\left(3+\frac{1}{n^{2}}\right)-\sqrt{\frac{1}{n}} \leqslant c\left(1-\frac{1}{n}\right)^{2},
$$

where $c<\frac{3}{4}, n \geqslant 1$.
Notice that as $n \rightarrow+\infty$, the left side of equation (4) $\rightarrow \frac{3}{4}$, and the right side of equation (4) $\rightarrow c$, leading to a contradiction.
Therefore, the minimum value of the real number $c$ is $\frac{3}{4}$.
(Xiao Enli, Li Chaohui, Shi Kejie, Fudan University High School, 200433)