A Hard Way And A Hard Way

Let our expression equal $k$

Since $a,b,c$ are positive real numbers, any combinations of sums, products, and quotients of $a,b,c$ will be a positive real number.

This means we can use the AM-GM Inequality on our expression.

First let's factor it as:

$\frac{9a(a+b)}{c(b+c)}+\frac{12b(b+c)}{a(a+c)}+\frac{16c(a+c)}{b(a+b)}=k$

AM-GM gives us:

$\frac{k}{3} \geq \sqrt[\huge 3]{\frac{(9)(a)(a+b)(12)(b)(b+c)(16)(c)(a+c)}{(c)(b+c)(a)(a+c)(b)(a+b)}}$

Everything under the radical cancels except for the $9 \times 12 \times 16$ , so we are left with

$\frac{k}{3} \geq \sqrt[\Large 3]{9 \times 12 \times 16}$ $\large k \geq 36$

We know by the AM-GM Inequality for positive real numbers $x,y,z$ that $\large \frac{x+y+z}{3} \geq \sqrt[3]{xyz}$ , and that equality is achieved if and only if $x=y=z$ .

So we now know that:

$\frac{9a(a+b)}{c(b+c)}=12, \frac{12b(b+c)}{a(a+c)}=12, \frac{16c(a+c)}{b(a+b)}=12$

Let's start with the middle equation, so we have:

$\frac{12b(b+c)}{a(a+c)}=12$ $b^{2}+bc=a^2+ac$

We can see by inspection that if $a=b$ , this equation holds true. Algebraically, we can simplify further:

$b^{2}-a^{2}=(a-b)c$ $-(b+a)=c$

We know that this case can't happen, since unless $a=b=c=0$ , either $b+a$ or $c$ is negative. $a,b,c>0$ , so none of these cases are within our constraints. This means equality will occur only when $a=b$ .

Looking back, we also know that:

$\frac{9a(a+b)}{c(b+c)}=12, \frac{16c(a+c)}{b(a+b)}=12$

Let's let $a=1$ , since we can later scale our answer without changing the ratio. This means that $b=1$ as well. In the first equation, we end up with:

$\frac{18}{c(1+c)}=12$

$c^{2}+c-\frac{3}{2}=0$

Using Quadratic Formula , we get:

$c=\frac{-1 \pm \sqrt{1-(4)(1)(\frac{-3}{2})}}{2}$

$c>0$ , so we know that the minus solution doesn't exist within our constraints.

$c=\frac{-1+\sqrt{7}}{2}$

To make sure this minimum is actually attained, we need to check the other equation as well:

$\frac{16c(a+c)}{b(a+b)}=12$

$a=b=1$ , so we have:

$\frac{16c(1+c)}{2}=12$

$c^{2}+c-\frac{3}{2}=0$

We could do quadratic formula again, but we can see that this is indeed the exact same quadratic as before.

Since we found positive real values of $a,b,c$ such that all of our previous conditions are satisfied, we can conclude that:

$a=1,b=1,c=\frac{-1+\sqrt{7}}{2}$

is a combination of $a,b,c$ that will give us our minimum. Thus, the ratio of $c:a$ is:

$\frac{-1+\sqrt{7}}{2}:1$

Expressed as a fraction is:

$\frac{-1+\sqrt{7}}{2}$

$-1,2,7$ are all integers and $7$ is not divisible by the square of any prime. Thus, our answer is:

$\large -1+2+7=\boxed{8}$