8 non-negative real numbers $a, b, c$ satisfy $a b + b c + c a = 1$. Find the minimum value of $\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a}$.
Proof 1 The minimum value sought is $\frac{5}{2}$.
Let $f(a, b, c)=\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}$, without loss of generality, assume $a \leqslant b \leqslant c$. We first prove
$$
f\left(0, a+b, c^{*}\right) \leqslant f(a, b, c),
$$

where $c^{*}=\frac{1}{a+b}, a b+b c+c a=1$.
In fact,
$$
\text { (1) } \begin{aligned}
& \Leftrightarrow \frac{1}{a+b}+\frac{1}{a+b+c^{*}}+\frac{1}{c^{*}} \leqslant \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \\
& \Leftrightarrow \frac{a+b+2 c^{*}}{\left(a+b+c^{*}\right) c^{*}} \leqslant \frac{a+b+2 c}{a b+b c+c a+c^{2}} \\
& \Leftrightarrow \frac{a+b+2 c^{*}}{1+c^{* 2}} \leqslant \frac{a+b+2 c}{1+c^{2}} \\
& \Leftrightarrow \frac{a+b+2 c^{*}}{a+b+2 c} \leqslant \frac{1+c^{* 2}}{1+c^{2}} \\
& \Leftrightarrow \frac{2\left(c^{*}-c\right)}{a+b+2 c} \leqslant \frac{\left(c^{*}-c\right)\left(c^{*}+c\right)}{1+c^{2}} \\
& \Leftrightarrow \frac{2}{a+b+2 c} \leqslant \frac{c^{*}+c}{1+c^{2}}
\end{aligned}
$$
(Here we use $c^{*}=\frac{1}{a+b}>\frac{1-a b}{a+b}=c$ )
$$
\begin{array}{l}
\Leftrightarrow 2+2 c^{2} \leqslant(a+b+2 c)\left(c^{*}+c\right) \\
\Leftrightarrow 2 \leqslant(a+b)\left(c^{*}+c\right)+2 c c^{*}=1+(a+b) c+2 c c^{*} \\
\Leftrightarrow 1 \leqslant 1-a b+\frac{2(1-a b)}{(a+b)^{2}} \\
\Leftrightarrow(a+b)^{2} a b \leqslant 2(1-a b) .
\end{array}
$$

Notice that $2(1-a b)=2 c(a+b) \geqslant \frac{2(a+b)^{2}}{2}$ (Here we use $c \geqslant b \geqslant$ $a$), thus to prove (2) holds, we only need to prove $(a+b)^{2} \geqslant(a+b)^{2} a b$, which only requires proving $1 \geqslant a b$. And this is obvious, thus (1) holds.
Using (1), we know $f(a, b, c) \geqslant \frac{1}{a+b}+\frac{1}{a+b+\frac{1}{a+b}}+(a+b)$.

Let $x=a+b+\frac{1}{a+b}$, then $f(a, b, c) \geqslant x+\frac{1}{x} \geqslant \frac{5}{2}$ (Here we use $x \geqslant$ 2, and $x+\frac{1}{x}$ is monotonically increasing when $x>1$). The equality is achieved when $a=0, b=c=1$. Therefore, the minimum value sought is $\frac{5}{2}$.

Proof 2 Clearly, when $a, b, c$ are all greater than $\frac{\sqrt{3}}{3}$, we have $a b+b c+c a>1$, which is not in line with the problem, so $a, b, c$ must have one that is not greater than $\frac{\sqrt{3}}{3}$. Without loss of generality, assume $b \leqslant \frac{\sqrt{3}}{3}$, then
$$
\begin{aligned}
S & =\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{a+2 b+c}{1+b^{2}}+\frac{1}{c+a} \\
& =\frac{2 b}{1+b^{2}}+\frac{a+c}{1+b^{2}}+\frac{1}{a+c} .
\end{aligned}
$$

Below, we fix $b$ and first find the minimum value of $\frac{a+c}{1+b^{2}}+\frac{1}{a+c}$.
For this, let $x=a+c$, then $f(x)=\frac{x}{1+b^{2}}+\frac{1}{x}$ is monotonically increasing when $x \geqslant \sqrt{1+b^{2}}$, and $x=a+c$, also
$$
1-b(a+c)=a c \leqslant \frac{(a+c)^{2}}{4},
$$

thus $\frac{x^{2}}{4} \geqslant 1-b x \Rightarrow x \geqslant \frac{-4 b+\sqrt{16 b^{2}+16}}{2}$. That is, $x \geqslant 2 \sqrt{b^{2}+1}-2 b$, combining $0 \leqslant b \leqslant \frac{\sqrt{3}}{3}$, we know $b^{2}+1 \geqslant 4 b^{2} \Rightarrow x \geqslant \sqrt{b^{2}+1}$, thus $f(x) \geqslant$ $f\left(2 \sqrt{b^{2}+1}-2 b\right)$.
In summary, we know:
$$
\begin{array}{l}
S \geqslant \frac{2 b}{1+b^{2}}+f\left(2 \sqrt{b^{2}+1}-2 b\right) \\
=\frac{2 b}{1+b^{2}}+\frac{2}{\sqrt{b^{2}+1}}-\frac{2 b}{1+b^{2}}+\frac{1}{2 \sqrt{b^{2}+1}-2 b} \\
\quad=\frac{2}{\sqrt{b^{2}+1}}+\frac{1}{2 \sqrt{b^{2}+1}-2 b}=\frac{2}{\sqrt{b^{2}+1}}+\frac{\sqrt{b^{2}+1}+b}{2} . \\
\text { Let } \sqrt{b^{2}+1}-b=y, \text { then } b^{2}+1=(y+b)^{2}=y^{2}+2 y b+b^{2} \Rightarrow \\
\frac{1-y^{2}}{} \text { (Notice }
\end{array}
$$
$b=\frac{1-y^{2}}{2 y}$ (Notice: $\sqrt{b^{2}+1}>b$, hence $y>0$). Thus
$$
\begin{aligned}
S & \geqslant \frac{2}{y+b}+\frac{1}{2 y}=\frac{2}{y+\frac{1-y^{2}}{2 y}}+\frac{1}{2 y} \\
& =\frac{4 y}{1+y^{2}}+\frac{1}{2 y}=\frac{8 y^{2}+1+y^{2}}{2 y\left(1+y^{2}\right)}=\frac{9 y^{2}+1}{2 y\left(1+y^{2}\right)} \\
& =\frac{5}{2}+\frac{9 y^{2}+1-5 y-5 y^{3}}{2 y\left(1+y^{2}\right)} \\
& =\frac{5}{2}+\frac{(1-y)\left(5 y^{2}-4 y+1\right)}{2 y\left(1+y^{2}\right)} \\
& =\frac{5}{2}+\frac{(1-y)\left(5\left(y-\frac{2}{5}\right)^{2}+\frac{1}{5}\right)}{2 y\left(1+y^{2}\right)} \geqslant \frac{5}{2},
\end{aligned}
$$

Here we use $\sqrt{b^{2}+1}-b \leqslant 1$.
The equality is achieved when $b=0$, at this time $a=c=1$. The minimum value sought is $\frac{5}{2}$.