Example 14 Let non-negative real numbers $a, b, c$ satisfy $ab + bc + ca = 1$. Find the minimum value of $\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a}$.
(2003 China National Team Training Problem)
Solution 1 It is evident that at least one of $a, b, c$ does not exceed $\frac{\sqrt{3}}{3}$. Without loss of generality, let $b \leqslant \frac{\sqrt{3}}{3}$, then
$$
S=\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{a+2 b+c}{1+b^{2}}+\frac{1}{c+a}=\frac{2 b}{1+b^{2}}+\frac{a+c}{1+b^{2}}+\frac{1}{a+c} .
$$

First, fix $b$ and find the minimum value of $\frac{a+c}{1+b^{2}}+\frac{1}{a+c}$.
Let $a+c=x$, then $f(x)=\frac{x}{1+b^{2}}+\frac{1}{x}$ is monotonically increasing when $x \geqslant \sqrt{1+b^{2}}$.
Also, $1-b(a+c)=a c \leqslant \frac{(a+c)^{2}}{4}$, which means $1-b x \leqslant \frac{x^{2}}{4}$. Therefore, $x \geqslant 2 \sqrt{1+b^{2}}-2 b$.
Considering $0 \leqslant b \leqslant \frac{\sqrt{3}}{3}$, we know $x \geqslant 2 \sqrt{b^{2}+1}-2 b \geqslant \sqrt{b^{2}+1}$,
thus $f(x) \geqslant f\left(2 \sqrt{b^{2}+1}-2 b\right)$.
In summary, $S \geqslant \frac{2 b}{1+b^{2}}+f\left(2 \sqrt{b^{2}+1}-2 b\right)=\frac{2 b}{1+b^{2}}+\frac{2}{\sqrt{1+b^{2}}}-\frac{2 b}{1+b^{2}}+$
$$
\frac{1}{2 \sqrt{b^{2}+1}-2 b}=\frac{2}{\sqrt{b^{2}+1}}+\frac{1}{2 \sqrt{b^{2}+1}-2 b}=\frac{2}{\sqrt{b^{2}+1}}+\frac{\sqrt{b^{2}+1}+b}{2} \text {. }
$$

Let $\sqrt{b^{2}+1}-b=y$, then $b^{2}+1=(y+b)^{2}$, solving this gives $b=\frac{1-y^{2}}{2 y}$.
Thus, $S \geqslant \frac{2}{y+b}+\frac{1}{2 y}=\frac{2}{y+\frac{1-y}{2 y}}+\frac{1}{2 y}=\frac{4 y}{1+y^{2}}+\frac{1}{2 y}$
$$
\begin{array}{l}
=\frac{9 y^{2}+1}{2 y\left(1+y^{2}\right)}=\frac{5}{2}+\frac{9 y^{2}+1-5 y-5 y^{3}}{2 y\left(1+y^{2}\right)}=\frac{5}{2}+\frac{(1-y)\left(5 y^{2}-4 y+1\right)}{2 y\left(1+y^{2}\right)} \\
=\frac{5}{2}+\frac{(1-y)\left[5\left(y-\frac{2}{5}\right)^{2}+\frac{1}{5}\right]}{2 y\left(1+y^{2}\right)} \geqslant \frac{5}{2} .
\end{array}
$$

The equality holds when $y=1$, i.e., $b=0$, at this time $a=c=1$, hence the minimum value sought is $\frac{5}{2}$.
Solution 2 Let $f(a, b, c)=\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}$, noting the condition, we know $a, b, c$ have symmetry, so we can assume $0 \leqslant a \leqslant b \leqslant c$.
First, prove $f\left(0, a+b, c^{\prime}\right) \leqslant f(a, b, c)$, where $c^{\prime}=\frac{1}{a+b}$.
That is, prove $\frac{1}{a+b}+\frac{1}{a+b+c^{\prime}}+\frac{1}{c^{\prime}} \leqslant \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}$.
Since $c^{\prime}=\frac{1}{a+b}, c=\frac{1-a b}{a+b}$, it is not difficult to verify that the above is equivalent to $(a+b)^{2} a b \leqslant 2(1-a b)$.
Noting that $2(1-a b)=2 c(a+b) \geqslant(a+b)^{2} \geqslant(a+b)^{2} \cdot a b$, hence (*) holds.
Thus, $f(a, b, c) \geqslant f\left(0, a+b, \frac{1}{a+b}\right)=\frac{1}{a+b}+\frac{1}{a+b+\frac{1}{a+b}}+a+b$.
Let $t=\frac{1}{a+b}+a+b$, then $t \geqslant 2$. Since the function $g(t)=t+\frac{1}{t}$ is increasing when $t \geqslant 2$, thus when $t \geqslant 2$, $g(t) \geqslant g(2)=\frac{5}{2}$, i.e., $\frac{1}{a+b}+\frac{1}{a+b+\frac{1}{a+b}}+a+b \geqslant \frac{5}{2}$, hence $f(a, b, c) \geqslant \frac{5}{2}$.