Minimizing Coefficients

With descartes rule we can say all roots are positive

Let the roots be $x, y, z$

$x + y + z = a$

$xy + yz + zx = b$

$xyz = 8$

Applying A.M-G.M

$\dfrac{x + y + z}{3} \geq \sqrt[3]{xyz}$

$x + y + z \geq 6$

$a_{min} = 6$

Now applying A.M -H.M

$\dfrac{ x + y + z}{3} \geq \dfrac{3xyz}{xy + yz + zx}$

Hence

$xy + yz + zx = b_{min} =\boxed{ 12}$

PS:- just apply G.M -H.M to find 'b'

It is easy to see that all the roots will be positive. Let the roots be $p$ , $q$ and $r$ . Now by Vieta,

$pq+qr+rs = b$

$pqr = 8$

Applying A.M-G.M $\frac{pq+qr+rs}{3} \geq (pqr)^{\frac{2}{3}}$ $\Rightarrow \frac{b}{3} \geq 8^{\frac{2}{3}}$ $\Rightarrow b \geq 12$ Hence answer is $\boxed{12}$

After looking at the equation for a while, I realize that it looked remarkable similar to the binomial expansion of (a-b)^3. After realizing this, the next logical step was just to reverse the formula for the expansion of a cubed binomial by taking 3rd root of the leading power and constant and ignoring the unknowns. This left me with (x-2)^3 which you can then expand back using the same binomial expansion formula to find the missing coefficients. After expanding you get x^3-6x^2+12x-8 which follows the pattern of the original equation. Given that all of the roots in (x-2)^3 are real, and a and b are positive real numbers, I decided I could reasonably assume that my coefficients were correct. b = 12 so 12 is your answer.