Stuck when transforming and solving this Given abc=1 ( all positive real numbers). Prove that:
$$\frac ab + \frac bc + \frac ca +3( \frac ba +\frac cb +\frac ac) \ge 2(a +b +c+\frac 1a+ \frac 1b +\frac1c)$$ 
My attempt:
$$\frac ab + \frac bc + \frac ca +3( \frac ba +\frac cb +\frac ac) \ge 2(a +b +c+\frac 1a+ \frac 1b +\frac1c)$$
Remove denominator
$$a^2c+ab^2+bc^2+3(b^2c+c^2a+a^2b) + a +b + c\ge 3(a+b+c) + 2(ab+bc+ca)$$
Turn this to
$$(a^2+1)c +(b^2+1)a + (c^2+1)b +3(b^2c+c^2a+a^2b)\ge 3(a+b+c) + 2(ab+bc+ca)$$ (1)
We have: $$(a^2+1)c \ge 2ac$$ (2)
From (1) and (2) i observe what is left to prove is $$3(b^2c+c^2a+a^2b)\ge 3(a+b+c)$$
Meaning to prove $$b^2c+c^2a+a^2b\ge a+b+c$$ given abc = 1. Which is where i stuck

The equality $a^2b+b^2c+c^2a\geq a+b+c$ we can prove by AM-GM:
$$\sum_{cyc}a^2b=\frac{1}{3}\sum_{cyc}(2a^2b+c^2a)\geq\frac{1}{3}\sum_{cyc}3\sqrt[3]{(a^2b)^2c^2a}=$$
$$=\sum_{cyc}\sqrt[3]{a^5b^2c^2}=\sum_{cyc}a.$$
Another way.
Let $a=\frac{x}{y}$ and $b=\frac {y}{z},$ where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{z}{x}$ and we need to prove that:
$$\sum_{cyc}\left(\frac{\frac{x}{y}}{\frac{y}{z}}+3\frac{\frac{y}{z}}{\frac{x}{y}}-\frac{2x}{y}-\frac{2y}{x}\right)\geq0$$ or
$$\sum_{cyc}(x^3y^3+3x^4yz-2x^3y^2z-2x^3z^2y)\geq0,$$ which is true by Muirhead or by SOS:
$$\sum_{cyc}(2x^3y^3-x^3y^2z-x^3z^2y+6x^4yz-3x^3y^2z-3x^3z^2y)\geq0$$ or
$$\sum_{cyc}(z^3(x^3-x^2y-xy^2+y^3)+3xyz(x^3-x^2y-xy^2+y^3))\geq0$$ or
$$\sum_{cyc}(x-y)^2(x+y)z(z^2+3xy)\geq0.$$
