Prove $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{a^{2} b+b^{2} c+c^{2} a}{a b^{2}+b c^{2}+c a^{2}} \geqslant \frac{5}{2}$
$$\begin{array}{l} \Leftrightarrow \sum_{\text {cyc }}\left(\frac{a}{b+c}-\frac{1}{2}\right) \geqslant \frac{\sum_{\text {cyc }}\left(a^{2} c-a^{2} b\right)}{\sum_{\text {cyc }} a^{2} c} \\ \Leftrightarrow \sum_{\text {cyc }} \frac{a-b-(c-a)}{2(b+c)} \geqslant \frac{(a-b)(b-c)(c-a)}{a^{2} c+b^{2} a+c^{2} b} \\ \Leftrightarrow \sum_{\text {cyc }} \frac{a-b}{2}\left(\frac{1}{b+c}-\frac{1}{a+c}\right) \geqslant \frac{(a-b)(b-c)(c-a)}{a^{2} c+b^{2} a+c^{2} b} \\ \quad \Leftrightarrow \sum_{\text {cyc }} \frac{(a-b)^{2}}{(a+c)(b+c)} \geqslant \frac{2(a-b)(b-c)(c-a)}{a^{2} c+b^{2} a+c^{2} b} \end{array}$$

If $(a-b)(b-c)(c-a) \leqslant 0$, then the original inequality is obviously true. Now assume $(a-b)(b-c)(c-a)>0$, and
$$\frac{a^{2} c+b^{2} a+c^{2} b}{a^{2} b+b^{2} c+c^{2} a}=t$$

Then $t>1$. By the Arithmetic Mean-Geometric Mean Inequality, we have
$$\sum_{c \text { yc }} \frac{(a-b)^{2}}{(a+c)(b+c)} \geqslant 3 \sqrt[3]{\frac{(a-b)^{2}(b-c)^{2}(c-a)^{2}}{(a+b)^{2}(a+c)^{2}(b+c)^{2}}}$$

Therefore, it suffices to prove
$$27\left(a^{2} c+b^{2} a+c^{2} b\right)^{3} \geqslant 8(a+b)^{2}(a+c)^{2}(b+c)^{2}(a-b)(b-c)(c-a)$$

Since
$$(a+b)(a+c)(b+c)=\sum_{\text {cyc }}\left(a^{2} b+a^{2} c\right)+2 a b c \leqslant \frac{4}{3} \sum_{\text {cyc }}\left(a^{2} b+a^{2} c\right)$$

It suffices to prove the following inequality:
$$27 t^{3} \geqslant 8 \cdot \frac{16}{9}(t+1)^{2}(t-1)$$

The above inequality is obviously true, hence the original inequality holds.