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

Question

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.

Answer 1

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

Answer 2

Alt. hint:  by AM-GM:

$$ \frac{1}{a}+\frac{1}{bc}=\frac{1}{a}+\frac{a+b+c}{bc} = \frac{1}{a} + \frac{1}{b}+\frac{1}{c}+\frac{a}{bc}\ge 4 \sqrt[4]{\frac{1}{b^2c^2}} = 4 \sqrt{\frac{1}{bc}} $$

Answer 3

Another way, is to use a more general Holder’s inequality: $$LHS \geqslant \left( \frac1{\sqrt[3]{abc}}+\frac1{\sqrt[3]{a^2b^2c^2}}\right)^3\geqslant (3+9)^3$$