For positive real numbers $\alpha, \beta, \gamma, x, y, z$, set $\alpha \beta \gamma = x y z = 1, x, y, z \in [\alpha, \gamma]$.
Since $(\alpha + \beta + \gamma) - (x + y + z)$
$$=\left\{\begin{array}{l} \beta \gamma(x - \alpha)(z - \alpha) + x z(\gamma - y)(y - \beta) \geqslant 0, \quad y \geqslant \beta, \\ \alpha \beta(\gamma - x)(\gamma - z) + x z(y - \alpha)(\beta - y) \geqslant 0, \quad y \leqslant \beta, \end{array}\right.$$

Therefore, $\square$
$$x + y + z \leqslant \alpha + \beta + \gamma$$

Define $\{\alpha, \beta, \gamma\} = \left\{\frac{b}{a}, \frac{c}{b}, \frac{a}{c}\right\}, \alpha \leqslant \beta \leqslant \gamma$. Note that $\frac{c + a}{b + c} \in \left[\min \left\{\frac{c}{b}, \frac{a}{c}\right\}, \max \left\{\frac{c}{b}, \frac{a}{c}\right\}\right] \subseteq [\alpha, \gamma]$, and similar conclusions hold for $\frac{a + b}{c + a}, \frac{b + c}{a + b}$, so
$$\frac{c + a}{b + c} + \frac{a + b}{c + a} + \frac{b + c}{a + b} \leqslant \alpha + \beta + \gamma = \frac{b}{a} + \frac{c}{b} + \frac{a}{c}.$$

Also, $\frac{b + a}{b + c} \in \left[\min \left\{\frac{a}{c}, 1\right\}, \max \left\{\frac{a}{c}, 1\right\}\right] \subseteq [\alpha, \gamma]$, and similar conclusions hold for $\frac{c + b}{c + a}, \frac{a + c}{a + b}$, so
$$\frac{b + a}{b + c} + \frac{c + b}{c + a} + \frac{a + c}{a + b} \leqslant \alpha + \beta + \gamma = \frac{b}{a} + \frac{c}{b} + \frac{a}{c}.$$

Adding (1) and (2), and using $a + b + c = 1$, we can rearrange to obtain the desired inequality.