Solution:
Denote the left side of the inequality by $LS$. By Hölder we have
$$\left(a\left(b^{2}+\lambda b c+c^{2}\right)+b\left(c^{2}+\lambda c a+a^{2}\right)+c\left(a^{2}+\lambda a b+b^{2}\right)\right)(LS)^{2} \geq (a+b+c)^{3}$$
So now it is sufficient to prove
$$\frac{(a+b+c)^{3}}{a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}+3 \lambda a b c} \geq \frac{9}{\lambda+2}$$
Now for easier notation write $\sum a^{2} b$ for $a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}$. Multiplying by the denominators and cancelling terms on both sides results in the inequality
$$2\left(a^{3}+b^{3}+c^{3}\right)+12 a b c+\lambda\left(a^{3}+b^{3}+c^{3}-21 a b c+3 \sum a^{2} b\right) \geq 3 \sum a^{2} b$$
Note that $a^{3}+b^{3}+c^{3}+3 \sum a^{2} b-21 a b c \geq 0$ by AM-GM. Since the $LS$ is a linear function in $\lambda$ and the coefficient for $\lambda$ is positive, the inequality is stricter for smaller $\lambda$. So in other words we can now assume $\lambda=1 / 4$. Multiplying by $4 / 9$ and rearranging once again we arrive at the final inequality:
$$\left(a^{3}+b^{3}+c^{3}\right)+3 a b c-\sum a^{2} b \geq 0$$
and this inequality is true by Schur.