Analysis: This problem requires appropriately transforming the inequality to be proven, making the forms on both sides symmetrical, to create conditions for using the rearrangement inequality.

Proof: The inequality to be proven is equivalent to
$$\frac{c^{2}}{a+b}+\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a} \geq \frac{a^{2}}{a+b}+\frac{b^{2}}{b+c}+\frac{c^{2}}{c+a},$$

By the cyclic symmetry of the inequality, without loss of generality, assume $a \leq b \leq c$, then
$$a^{2} \leq b^{2} \leq c^{2}, \quad a+b \leq a+c \leq b+c$$
$$\frac{1}{a+b} \geq \frac{1}{a+c} \geq \frac{1}{b+c}, \quad \frac{c^{2}}{a+b}+\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a} \text{ is the sum in ascending order, } \frac{a^{2}}{a+b}+\frac{b^{2}}{b+c}+\frac{c^{2}}{c+a} \text{ is the sum in random order,}$$
$$\therefore \frac{c^{2}}{a+b}+\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a} \geq \frac{a^{2}}{a+b}+\frac{b^{2}}{b+c}+\frac{c^{2}}{c+a},$$

which means $\frac{c^{2}-a^{2}}{a+b}+\frac{a^{2}-b^{2}}{b+c}+\frac{b^{2}-c^{2}}{c+a} \geq 0$.