Prove $\left(\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b}\right)\left(\frac{ab}{c^2+b^2} + \frac{bc}{a^2 + c^2} + \frac{ca}{a^2 + b^2}\right) \ge 9$ If $a,b,c \in \mathbb{R^+}$,then prove that the following inequality holds:
$$\left(\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b}\right)\left(\frac{ab}{c^2+b^2} + \frac{bc}{a^2 + c^2} + \frac{ca}{a^2 + b^2}\right) \ge 9 \tag{1}$$

This is a inequality posted on the AoPS site and it's been answered by one user. The problem is that I can't quite understand the proof. The proof is:
After using AM-GM we need to prove:
$$(a^2c + a^2b + b^2c + b^2a + c^2a + c^2b)^3 \ge 27abc(a^2+b^2)(a^2+c^2)(b^2+c^2) \tag{2}$$
which holds from AM-GM.
The second part is clear to me, but I can't understand the part when he transforms $(1)$ into $(2)$. Because he said he used AM-GM I apply that inequality on both terms in $(1)$, but i end up with:
$$\sqrt[3]{\frac{(a+b)(b+c)(c+a)a^2b^2c^2}{abc(a^2+b^2)(a^2+c^2)(b^2+c^2)}} \ge 1$$
And I don't know how to continue. I used the fact that:$$(a+b)(b+c)(c+a) = a^2c + a^2b + b^2c + b^2a + c^2a + c^2b + 2abc$$
But the term $a^2b^2c^2$ in the numerator is making troubles.
From AM-GM  and Nesbitt's Inequality we know that the first term on the LHS in $(1)$ is $\ge 6$ and i tried proving 
$$\frac{ab}{c^2+b^2} + \frac{bc}{a^2 + c^2} + \frac{ca}{a^2 + b^2} \ge \frac 32$$
But again I end up with empty hands.
How can I prove it? Can someone explain me the first part in the AoPS answer?

As njguliyev pointed out,
$$\left(\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b}\right) = \frac{a^2c + a^2b + b^2c + b^2a + c^2a + c^2b}{abc}$$
$$\left(\frac{ab}{c^2+b^2} + \frac{bc}{a^2 + c^2} + \frac{ca}{a^2 + b^2}\right) \ge 3 \frac{\sqrt[3]{(abc)^2}}{\sqrt[3]{(a^2+b^2)(a^2+c^2)(b^2+c^2)}}$$
