By the Cauchy-Buniakowski inequality,
$$(\sqrt{a}^2 + \sqrt{b}^2 + \sqrt{c}^2)((\sqrt{a}(b+c))^2 + (\sqrt{b}(c+a))^2 + (\sqrt{c}(a+b))^2) \geq (a(b+c) + b(c+a) + c(a+b))^2 = 4(ab + bc + ca)^2.$$
Hence,
$$(a+b+c)(a(b+c)^2+b(c+a)^2+c(a+b)^2) \geq 4(ab+bc+ca)^2 \geq 4(ab+bc+ca)(a+b+c) \Leftrightarrow (a(b+c)^2+b(c+a)^2+c(a+b)^2) \geq 4(ab+bc+ca).$$
It remains to note that
$$a(b+c)^2 + b(c+a)^2 + c(a+b)^2 = (a+b+c)(ab+bc+ca) + 3abc,$$
which gives the required inequality.

Alternative solution:

Let $\sigma_1 = a+b+c$, $\sigma_2 = ab+bc+ca$, $\sigma_3 = abc$. By the problem condition, $\sigma_2 \ge \sigma_1$. Since $\sigma_1^2 \ge 3\sigma_2$ (well-known inequality), we have $\sigma_1^2 \ge 3\sigma_2 \ge 3\sigma_1$, i.e. $\sigma_1 \ge 3$.
Now we use Schur's inequality $\sigma_1^3 + 9\sigma_3 \ge 4\sigma_1\sigma_2$. We have $\sigma_1^2\sigma_2 \ge \sigma_1^3$ and $3\sigma_1\sigma_3 \ge 9\sigma_3$, hence $\sigma_1^2\sigma_2 + 3\sigma_1\sigma_3 \ge 4\sigma_1\sigma_2$ and then $\sigma_1\sigma_2 + 3\sigma_3 \ge 4\sigma_2$, as required.

Alternative solution:

Since $ab+bc+ca \ge a+b+c$, it suffices to prove the inequality $(a+b+c)^2(ab+bc+ca) + 3abc(a+b+c) \ge 4(ab+bc+ca)^2$. The last inequality can be easily transformed to
$$a^3b + b^3a + a^3c + c^3a + b^3c + c^3b \ge 2(a^2b^2 + b^2c^2 + c^2a^2),$$
which is true by Muirhead's theorem.