Inequality involving AM-GM but its wierd Let a, b, c be positive real numbers. Prove that
$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+ \frac{3\cdot\sqrt[3]{abc}}{a+b+c} \geq 4$
Ohk now i know using AM-GM that 
$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq 3 \cdot \sqrt[3] {\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a}} = 3 \cdot 1=3 $
Now if i could have shown that the other term $\geq 1$. I would be done.
But the problem is that (again using AM-GM)
$\sqrt[3]{abc} \leq \frac{a+b+c}{3}
\Rightarrow 3 \cdot \sqrt[3]{abc} \leq a+b+c \Rightarrow \frac{3 \cdot \sqrt[3]{abc}}{a+b+c} \leq 1$
So if first part is $\geq 3$ and second part is $\leq 1$, How will i show that it is greater than $4$?
Is my approach correct? Or is there something wrong with the question?
Thanks.
(Source: https://web.williams.edu/Mathematics/sjmiller/public_html/161/articles/Riasat_BasicsOlympiadInequalities.pdf ,pg-$13$ exercise $1.3.4.a$ )

From
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant \frac{(a+b+c)^2}{ab+bc+ca},$$
and $ \sqrt[3]{abc} \geqslant \frac{3abc}{ab+bc+ca},$ we need to prove
$$\frac{(a+b+c)^2}{ab+bc+ca} + \frac{9abc}{(ab+bc+ca)(a+b+c)} \geqslant 4,$$
equivalent to
$$ (a+b+c)^2 + \frac{9abc}{a+b+c} \geqslant 4(ab+bc+ca),$$
or
$$a^2+b^2+c^2 + \frac{9abc}{a+b+c} \geqslant 2(ab+bc+ca).$$
Which is Schur is inequality.
