Alternative Solution by PSC. By the Power Mean Inequality we have

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

So it is enough to prove that

$$(a+b+c)^{2}\left(\frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a}\right) \geqslant 9$$

or equivalently, that

$$(a+b+c)^{2}\left(\frac{1}{a c+b c}+\frac{1}{b a+c a}+\frac{1}{c b+a b}\right) \geqslant \frac{27}{2}$$

Since $(a+b+c)^{2} \geqslant 3(a b+b c+c a)=\frac{3}{2}((a c+b c)+(b a+c a)+(c b+a c))$, then (5) follows by the Cauchy-Schwarz Inequality.