Example 2 Let $a, b, c \in \mathbf{R}$ and $a+b+c=1$. Find the value of $\min \{\max \{a+b, b+c, c+a\}\}$.
(2001, Beijing High School Mathematics Competition)
Solution: Let $x=\max \{a+b, b+c, c+a\}$. Then
$$
\begin{array}{l}
x=\frac{x+x+x}{3} \\
\geqslant \frac{(a+b)+(b+c)+(c+a)}{3} \\
=\frac{2}{3}(a+b+c)=\frac{2}{3} .
\end{array}
$$

When $a+b=b+c=c+a=x=\frac{2}{3}$, we can solve for $a=b=c=\frac{1}{3}$ and $x=\frac{2}{3}$. Therefore,
$$
\min \{\max \{a+b, b+c, c+a\}\}=\frac{2}{3} \text {. }
$$