Proving $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$ for positive $a$, $b$, $c$ 
For $a,b,c>0$ Prove that $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$$

My attempt: By AM-GM we obtain $$\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\ge 3\sqrt[3]{\frac{a^2}{bc}}=\frac{3a}{\sqrt[3]{abc}}$$
Thus $$\sum \frac{a+c}{b}\ge \frac{2(a+b+c)}{\sqrt[3]{abc}}$$
So it suffices to show that $$6\ge \frac{2(a+b+c)}{\sqrt[3]{abc}}\Leftrightarrow 3\sqrt[3]{abc}\ge a+b+c$$
Which is clearly wrong. :"(
Thank you very much.

Due to homogeneity, assume that $abc = 1$.
Let $p = a+b+c, q = ab+bc+ca, r=abc=1$.
We need to prove that $p\cdot \frac{q}{r} + 3 \ge 4 \cdot \frac{p}{\sqrt[3]{r}}$
or $pq + 3 \ge 4 p$.
Since $q^2 \ge 3pr$, it suffices to prove that
$p \cdot \sqrt{3p} + 3 \ge 4p$ or $\frac{1}{3}(\sqrt{3p} - 3)(3p - \sqrt{3p} - 3)\ge 0$
which is true since $p\ge 3$. We are done.
