Show that the following inequality is true: $(\frac{1}{a} + \frac{1}{bc}) (\frac{1}{b} + \frac{1}{ca})(\frac{1}{c} + \frac{1}{ab}) \geq 1728$ This is a question from a past Olympiad paper:

Three positive real numbers $a, b, c$ satisfy the following constraint:
  $a+b+c = 1$. Show that the following inequality is true:
$\left(\dfrac{1}{a} + \dfrac{1}{bc}\right) \left(\dfrac{1}{b} + \dfrac{1}{ca}\right)\left(\dfrac{1}{c} + \dfrac{1}{ab}\right) \geq 1728$.

Starting from the L.H.S, I end up with:
$\dfrac{abc(abc+a^2 + b^2 + c^2 + 1) + a^2b^2 + a^2c^2+b^2c^2}{(abc)^3}$.
From the numerator I suspect $a+b+c$ will be factorised, but this is the furthest I have got to.
Using the hint given by timon92:
$\dfrac{1}{a} + \dfrac{1}{bc} = \dfrac{(a+b)(a+c)}{abc}$
and likewise for other two, I end up with:
$(a+b)^2(b+c)^2(a+c)^2 \geq 1728(abc)^3$
$\left((a+b)(b+c)(a+c)\right)^\frac{2}{3} \geq 8abc$
Many thanks in advance.

Hint: $$\frac 1a + \frac{1}{bc} = \frac{bc+a}{abc} = \frac{bc+a(a+b+c)}{abc} = \frac{(a+b)(a+c)}{abc}.$$
