Example 11 Given that $A A_{1}, B B_{1}, C C_{1}$ are the angle bisectors of $\triangle A B C$, and $A_{1}, B_{1}, C_{1}$ are on sides $B C, C A, A B$ respectively. Let $B C=a, C A=b, A B=c$. If $A_{1}, B_{1}, C_{1}, B$ are concyclic, find the minimum value of $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$.
Solve as shown in Figure 3-4, let the circle determined by points $A_{1}, B_{1}, C_{1}, B$ intersect line $AC$ at point $D$. Since points $A, C$ are outside the circle, $B_{1}$ is on the circle, and $B_{1}$ is on side $AC$, thus $D$ must also be on side $AC$.
By the secant theorem, we have $A C_{1} \cdot A B = A D \cdot A B_{1}$.
Also, by the angle bisector property, we have $A C_{1} = (A C_{1} + C_{1} B) \cdot \frac{A C}{A C + C B} = \frac{b c}{a + b}, A B_{1} = (A B_{1} + B_{1} C) \cdot \frac{A B}{A B + B C} = \frac{b c}{c + a}$.
Therefore, $A D = \frac{A C_{1} \cdot A B}{A B_{1}} = \frac{b c}{a + b} \cdot c \cdot \frac{c + a}{b c} = (a + c) \cdot \frac{c}{a + b}$.
Similarly, $C D = (a + c) \cdot \frac{a}{b + c}$.
Thus, $b = A C = A D + D C = (a + c) \left( \frac{c}{a + b} + \frac{a}{b + c} \right)$,
which means $\frac{b}{a + c} = \frac{c}{a + b} + \frac{a}{b + c}$.
Therefore, the model $\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b}$ transforms into $\frac{2 b}{c + a}$.
Let $a + b = x, b + c = y, c + a = z$, then $a + b + c = \frac{1}{2}(x + y + z)$, which gives $a = \frac{x + z - y}{2}, b = \frac{x + y - z}{2}, c = \frac{y + z - x}{2}$. At this point, by transforming equation (*), we get
$$
\begin{aligned}
\frac{x + y}{z} & = \frac{y + z}{x} + \frac{x + z}{y} - 1 = \frac{z}{x} + \frac{z}{y} + \frac{y}{x} + \frac{x}{y} - 1 \\
& \geqslant \frac{z}{x} + \frac{z}{y} + 1 \geqslant z \cdot \frac{4}{x + y} + 1,
\end{aligned}
$$

where the equality holds if and only if $x = y$.
Let $\frac{x + y}{z} = t$, then the above inequality transforms into $t \geqslant \frac{4}{t} + 1$.
Solving the above inequality, we get $t \geqslant \frac{1}{2}(1 + \sqrt{17})$ or $t \leqslant \frac{1}{2}(1 - \sqrt{17})$ (discarded).
Thus, $\frac{b}{a + c} = \frac{x + y - z}{2 z} = \frac{t}{2} - \frac{1}{2} \geqslant \frac{1}{4}(\sqrt{17} - 1)$.
Therefore, $\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} = \frac{2 b}{a + c} \geqslant \frac{1}{2}(\sqrt{17} - 1)$.

The equality holds if and only if $x = y$, which implies $a = c$.