One, (20 points) Given that $a$, $b$, and $c$ are positive real numbers, and $abc=1$. Find the minimum value of
$$
\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}
$$
When $a=b=c=1$, the original expression $=\frac{3}{4}$.
Below, we compare the original expression with $\frac{3}{4}$.
$$
\begin{array}{l}
\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+ \\
\frac{c}{(c+1)(a+1)}-\frac{3}{4} \\
=\frac{a b+a c+b c+a+b+c-6}{4(a+1)(b+1)(c+1)} .
\end{array}
$$

Since $a b c=1$, and $a, b, c$ are positive real numbers, then
$$
\begin{array}{l}
a b+a c+b c+a+b+c-6 \\
=\left(a+\frac{1}{a}-2\right)+\left(b+\frac{1}{b}-2\right)+\left(c+\frac{1}{c}-2\right) \\
\geqslant 0 .
\end{array}
$$

Also, $4(a+1)(b+1)(c+1)>0$, hence
$$
\begin{array}{l}
\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)} . \\
\geqslant \frac{3}{4} .
\end{array}
$$

Therefore, the minimum value of the given algebraic expression is $\frac{3}{4}$.