Inequality. $(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a)$ Let $a,b,c$ be three real positive(strictly) numbers. Prove that: 
$$(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a).$$
I tried : 
$$abc\left(a+\frac{bc}{a}\right)\left(b+\frac{ca}{b}\right)\left(c+\frac{ab}{c}\right)\geq abc(a+b)(b+c)(c+a) $$
and now I want to try to prove that for example $$a+\frac{bc}{a} \geq a+b$$
but I don't know if is is a good idea. 
Thanks:) 

Here $\prod_{cyc}$ refers to the cyclic product of $x, y, z$.
Let $a=x^2, b=y^2, c=z^2$ for positives $x, y, z$. Then by the Cauchy-Schwarz inequality, we have:
$$LHS^2=\prod_{cyc}[(x^4+y^2z^2)(x^2z^2+y^4)]\ge \prod_{cyc}(x^3z+y^3z)^2=x^2y^2z^2\prod_{cyc}(x^3+y^3)^2$$
Then, by Power-Mean and AM-GM:
$$x^2y^2z^2\prod_{cyc}(x^3+y^3)^2\ge x^2y^2z^2\prod_{cyc}[\frac{(x^2+y^2)^3}{2}]\ge x^2y^2z^2\prod_{cyc}[(x^2+y^2)^2xy]=RHS^2$$
Thus $LHS^2\ge RHS^2$, and both sides are positive so $LHS\ge RHS$ as desired.
Sidenote: There is also a proof by direct expansion in $a, b, c$: upon expanding and rearranging the inequality becomes $\sum_{cyc}(a-b)^2(\frac{a+b}{2})(c^3+abc)\ge 0$, which is clear.
