$ 27 + ( 2 + \frac{a^{2}}{bc} ) (2 + \frac{b^{2}}{ac}) (2 + \frac{c^{2}}{ab}) \ge 6 (a+b+c)(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) $ Let $a,b,c$ be positive numbers. Prove that 
$$ 27 + ( 2 + \frac{a^{2}}{bc} ) (2 + \frac{b^{2}}{ac}) (2 + \frac{c^{2}}{ab}) \ge  6 (a+b+c)(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) $$

Attempt:
The inequality is equivalent to
$$ 36 + 4 \left( \frac{ab}{c^{2}} + \frac{ac}{b^{2}} + \frac{bc}{a^{2}} \right) + 2 \left( \frac{c^{2}}{ab} + \frac{b^{2}}{ac} + \frac{a^{2}}{bc}  \right) $$
$$ \ge 6 \left( 3 + \frac{a}{b} + \frac{b}{c} + \frac{c}{a} + \frac{b}{a} + \frac{c}{b} + \frac{a}{c} \right) $$
I have tried using $HM-GM$ but still does not get the result. Rewriting the above 
$$ \underbrace{1+1+...+1}_{18} + \underbrace{ \frac{ab}{c^{2}} + \frac{ac}{b^{2}} + \frac{bc}{a^{2}} + ... +  \frac{ab}{c^{2}} + \frac{ac}{b^{2}} + \frac{bc}{a^{2}} }_{4} + \left( \frac{c^{2}}{ab} + \frac{b^{2}}{ac} + \frac{a^{2}}{bc}  + \frac{c^{2}}{ab} + \frac{b^{2}}{ac} + \frac{a^{2}}{bc}  \right) $$
$$ \ge \underbrace{\left( \frac{a}{b} + \frac{b}{c} + \frac{c}{a} + \frac{b}{a} + \frac{c}{b} + \frac{a}{c} \right) + ... + \left( \frac{a}{b} + \frac{b}{c} + \frac{c}{a} + \frac{b}{a} + \frac{c}{b} + \frac{a}{c} \right) }_{6} $$
The left hand side and right hand side each has $36$ terms.

A full expanding gives:
$$\sum_{cyc}(a^3b^3+2a^4bc-3a^3b^2c-3a^3c^2b+3a^2b^2c^2)\geq0$$ or
$$\sum_{cyc}(a^3b^3-a^3b^2c-a^3c^2b+a^2b^2c^2)+abc\sum_{cyc}(a^3-a^2b-a^2c+abc)\geq0,$$
which true by Schur twice.
