Example 5 Find the real number triples $(a, b, c)$ satisfying $0<a, b, c \leqslant 1$, such that
$$
\begin{array}{l}
\min \left\{\sqrt{\frac{a b+1}{a b c}}, \sqrt{\frac{b c+1}{a b c}}, \sqrt{\frac{c a+1}{a b c}}\right\} \\
=\sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}} .
\end{array}
$$
By symmetry, without loss of generality, assume $a \geqslant b \geqslant c$.
Notice that, $00$.
Then $\sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}=u+v+w$.
Since $a \geqslant b \geqslant c \geqslant 0$, it follows that $a b \geqslant c a \geqslant b c$.
Thus, $\min \left\{\sqrt{\frac{a b+1}{a b c}}, \sqrt{\frac{b c+1}{a b c}}, \sqrt{\frac{c a+1}{a b c}}\right\}$
$=\sqrt{\frac{b c+1}{a b c}}$
$$
=\sqrt{\left(1+u^{2}\right)\left(1+\left(1+v^{2}\right)\left(1+w^{2}\right)\right)}.
$$

Notice that,
$$
\begin{array}{l}
\left(1+u^{2}\right)\left(1+\left(1+v^{2}\right)\left(1+w^{2}\right)\right) \\
=\left(1+u^{2}\right)\left(1+1+v^{2} w^{2}+v^{2}+w^{2}\right) \\
\geqslant\left(1+u^{2}\right)\left(1+2 v w+v^{2}+w^{2}\right) \\
=\left(1+u^{2}\right)\left((v+w)^{2}+1\right) \\
\geqslant(v+w+u)^{2} \text{ (Cauchy-Schwarz inequality), }
\end{array}
$$

Equality holds if and only if $v w=1, u(v+w)=1$.
$$
\begin{array}{l}
\text{Then } a=\frac{1}{1+\left(w+\frac{1}{w}\right)^{2}}=\frac{w^{2}}{1+\left(w^{2}+1\right)^{2}}, \\
b=\frac{1}{1+\left(\frac{1}{w}\right)^{2}}=\frac{w^{2}}{1+w^{2}}, c=\frac{1}{1+w^{2}},
\end{array}
$$

where $w \in \mathbf{R}_{+}$.
In summary,
$(a, b, c)$
$$
=\left(\frac{w^{2}}{1+\left(w^{2}+1\right)^{2}}, \frac{w^{2}}{1+w^{2}}, \frac{1}{1+w^{2}}\right)(w>0).
$$