One, (20 points) When $1<x<5$, find the minimum value of the algebraic expression
$$
\frac{\sqrt{(7 x+1)(5-x)(x-1)(7 x+5)}}{x^{2}}
$$
$$
\begin{array}{l}
\frac{\sqrt{(7 x+1)(5-x)(x-1)(7 x+5)}}{x^{2}} \\
=\sqrt{\frac{\left(7 x^{2}-2 x-5\right)\left(-7 x^{2}+34 x+5\right)}{x^{4}}} \\
=\sqrt{\left(7 x-\frac{5}{x}-2\right)\left(-7 x+\frac{5}{x}+34\right)} \text {. } \\
\end{array}
$$

Let $t=7 x-\frac{5}{x}$. Then
$$
\begin{array}{l}
\text { Equation (1) }=\sqrt{(t-2)(34-t)} \\
=\sqrt{-t^{2}+36 t-68} \\
=\sqrt{-(t-18)^{2}+256} \leqslant 16 .
\end{array}
$$

When $t=7 x-\frac{5}{x}=18$, i.e., $x=\frac{9+2 \sqrt{29}}{7}$,
the expression reaches its maximum value of 16.