10 non-negative real numbers $a, b, c$ satisfy: $a b + b c + a = 1$, find the minimum value of $\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a}$.
10. Let $f(a, b, c)=\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}$, and assume without loss of generality that $a \leqslant b \leqslant c$. We first prove: $f\left(0, a+b, c^{\prime}\right) \leqslant f(a, b, c)$. Here $c^{\prime}=\frac{1}{a+b}, a b+b c+c a=1$. In fact, $f\left(0, a+b, c^{\prime}\right) \leqslant f(a, b, c)$ is equivalent to
$$\frac{1}{a+b}+\frac{1}{a+b+c^{\prime}}+\frac{1}{c^{\prime}} \leqslant \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}$$

Since $c^{\prime}=\frac{1}{a+b}, \frac{1-a b}{a+b}=c$, it is not difficult to simplify the above to
$$(a+b)^{2} a b \leqslant 2(1-a b).$$

Notice that $2(1-a b)=2 c(a+b) \geqslant \frac{2(a+b)^{2}}{2} \geqslant(a+b)^{2} a b$, hence the conclusion holds.
Therefore, $f(a, b, c) \geqslant \frac{1}{a+b}+\frac{1}{a+b+\frac{1}{a+b}}+a+b$.
It is not difficult to verify that when $a=0, b=c=1$, $f$ achieves its minimum value $\frac{5}{2}$.
Note: The method of repeated extremum can also be used to solve:
Clearly, when $a, b, c$ are all greater than $\frac{\sqrt{3}}{3}$, we have $a b+b c+c a>1$, which contradicts the condition, so at least one of $a, b, c$ must be no 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 first fix $b$ and find the minimum value of $\frac{a+c}{1+b^{2}}+\frac{1}{a+c}$.
Let $x=a+c$, then $f(x)=\frac{x}{1+b^{2}}+\frac{1}{x}$ is monotonically increasing for $x \geqslant \sqrt{1+b^{2}}$, and $1-$ $b(a+c)=a c \leqslant \frac{(a+c)^{2}}{4}$, hence $\frac{x^{2}}{4} \geqslant 1-b x$, then $x \geqslant 2 \sqrt{b^{2}+1}-2 b$. Also, $b^{2}+1 \geqslant$ $4 b^{2}$, so $x \geqslant \sqrt{b^{2}+1}$, thus $f(x) \geqslant f\left(2 \sqrt{b^{2}+1}-2 b\right)$.

In summary, $S \geqslant \frac{2 b}{1+b^{2}}+f\left(2 \sqrt{b^{2}+1}-2 b\right)=\frac{2}{\sqrt{b^{2}+1}}+\frac{\sqrt{b^{2}+1}+b}{2}$, let $y=\sqrt{b^{2}+1}-b>0$, then $b=\frac{1-y^{2}}{2 y}$.
Thus, $S \geqslant \frac{2}{y+b}+\frac{1}{2 y}=\frac{9 y^{2}+1}{2 y\left(1+y^{2}\right)}=\frac{5}{2}+\frac{(1-y)\left(5\left(y-\frac{2}{5}\right)^{2}+\frac{1}{5}\right)}{2 y\left(1+y^{2}\right)}$ $\frac{5}{2}$ (note $\sqrt{b^{2}+1}-b \leqslant 1$ ).

The equality holds when $b=0$, at this time $a=c=1$, hence the minimum value sought is $\frac{5}{2}$.