Prove $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} + \frac{3\sqrt[3]{abc}}{a+b+c} \geq 4$

Question

$$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} + \frac{3\sqrt[3]{abc}}{a+b+c} \geq 4$$
My working out:
By AM-GM; $$\frac{a}{b} + \frac{a}{b} + \frac bc \geq 3\sqrt[3]\frac{a^2}{bc} = 3\frac{a}{\sqrt[3]{abc}}$$
Repeating this on $$\frac bc \quad and \quad \frac ca$$ we get the following: $$\frac ab +\frac bc + \frac ca \geq \frac {a + b + c}{\sqrt[3]{abc}}$$ Hence, $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} + \frac{3\sqrt[3]{abc}}{a+b+c} \geq \frac {a + b + c}{\sqrt[3]{abc}} + \frac{3\sqrt[3]{abc}}{a+b+c} \geq 2\sqrt3$$ I also tried minimising both parts of LHS, as the following:
$$\frac ab + \frac bc + \frac ca \geq 3$$ $$a + b + c \geq 3\sqrt[3]{abc} \implies \frac {\sqrt[3]{abc}}{a+b+c} \leq \frac {1} {3}$$ Hence, the sum of the minimum of both parts is $$3 + 3(\frac 13) = 4$$ occurring when a = b = c. However, this is not sufficient to prove the statement as we do not consider how each part changes in relationship to the other.

It would be much appreciated if someone can point me in the right direction? Thanks