By the Cauchy-Bunyakovskii inequality, we have
$$(a^2b + b^2c + c^2a)(a+b+c) \ge (ab + bc + ca)^2. \quad (*)$$

Now consider the given equality. Multiplying both its sides by $(a+b+c)abc$, we get
$$A = (a+b+c)\left(\frac{a^2bc}{b+c} + \frac{b^2ca}{c+a} + \frac{c^2ab}{a+b}\right) = abc(a+b+c) + \frac{(a+b+c)(a^2b + b^2c + c^2a)}{6} \geq \text{[see } (*)\text{]} \geq abc(a+b+c) + \frac{1}{6}(ab+bc+ca)^2.$$
Further,
$$A = \sum_{cyc} \left(\frac{a^3bc}{b+c} + a^2bc\right) = \sum_{cyc} \frac{a^3bc}{b+c} + abc(a+b+c) \geq abc(a+b+c) = \frac{1}{6}(ab+bc+ca)^2,$$
whence the required inequality follows.