Example 2 Let non-negative real numbers $a, b, c$ satisfy $ab + bc + ca = 1$. Find the minimum value of $u = \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a}$.
Given:
$$
(b+c)(c+a)=1+c^{2} \text {. }
$$

Taking the reciprocal and transforming, we get
$$
\frac{1}{b+c}+\frac{1}{c+a}=\frac{2 c+(a+b)}{1+c^{2}} \text {. }
$$

Thus, \( u=\frac{1}{a+b}+\frac{a+b}{1+c^{2}}+\frac{2 c}{1+c^{2}} \).
From \( a b+b c+c a=1 \),
$$
\begin{array}{l}
\Rightarrow \frac{1}{4}(a+b)^{2}+c(a+b) \geqslant 1 \\
\Rightarrow a+b \geqslant 2\left(\sqrt{1+c^{2}}-c\right) .
\end{array}
$$

By symmetry, without loss of generality, let \( a \geqslant b \geqslant c \), then
$$
0 \leqslant c \leqslant \frac{\sqrt{3}}{3}, \sqrt{1+c^{2}}-2 c \geqslant 0 \text {. }
$$

Let \( t=a+b \), then \( t \in\left[2 \sqrt{1+c^{2}}-2 c,+\infty\right) \).
Thus, \( u=\frac{1}{t}+\frac{t}{1+c^{2}}+\frac{2 c}{1+c^{2}} \).
It is easy to prove that \( u(t) \) is increasing on \( \left[2 \sqrt{1+c^{2}}-2 c,+\infty\right) \). Therefore,
$$
\begin{array}{l}
u(t) \geqslant u\left(2 \sqrt{1+c^{2}}-2 c\right) \\
=\frac{1}{2}\left(\sqrt{1+c^{2}}+c\right)+\frac{2}{\sqrt{1+c^{2}}} \\
=2\left(\sqrt{1+c^{2}}+\frac{1}{\sqrt{1+c^{2}}}\right)-\frac{3}{2}\left(\sqrt{1+c^{2}}-\frac{1}{3} c\right) \\
\geqslant 4-\frac{3}{2}\left(\sqrt{1+c^{2}}-\frac{1}{3} c\right) .
\end{array}
$$

Since \( \sqrt{1+c^{2}}+\frac{1}{\sqrt{1+c^{2}}} \geqslant 2 \), with equality if and only if \( c=0 \). And when \( 0 \leqslant c \leqslant \frac{\sqrt{3}}{3} \), we have
$$
g(c)=-\left(\sqrt{1+c^{2}}-\frac{1}{3} c\right) \geqslant g(0)=-1 \text {, }
$$

Therefore, \( u \geqslant \frac{5}{2} \).
Equality holds if and only if \( c=0, a=b=1 \).
Thus, \( u_{\text {min }}=\frac{5}{2} \).
Note: The key to this solution is to convert the problem into studying the monotonicity of the function
$$
u=\frac{1}{t}+\frac{t}{1+c^{2}}+\frac{2 c}{1+c^{2}}
$$

on \( \left[2 \sqrt{1+c^{2}}-2 c,+\infty\right) \).