By the AH mean inequality, we have

$$\frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a}=\frac{2}{3(a c+b c)}+\frac{2}{3(a b+a c)}+\frac{2}{3(a b+a c)} \geqslant \frac{3}{a b+a c+b c}$$

so it only remains to prove that $\frac{3}{a b+a c+b c} \geqslant \frac{a+b+c}{a^{3}+b^{3}+c^{3}}$, or equivalently

$$3\left(a^{3}+b^{3}+c^{3}\right) \geqslant(a+b+c)(a b+a c+b c)$$

The last inequality easily follows by summing $a^{3}+b^{3} \geqslant a b(a+b), a^{3}+c^{3} \geqslant a c(a+c)$, $b^{3}+c^{3} \geqslant b c(b+c)$ and $a^{3}+b^{3}+c^{3} \geqslant 3 a b c$.