$$ \begin{array}{l}\text { The minimum value of ab if roots of the equation } x^{3}-a x^{2}+b x-2=0 \\ \text { are positive, is }\end{array} $$
$$ \begin{array}{l}\text { Let } f(x)=x^{3}-a x^{2}+b x-2 \\ \therefore f^{\prime }(x)=3 x^{2}-2 a x+b \\ x_{-}=\frac{2 a-\sqrt{4 a^{2}-12 b}}{6} \\ \text { Also } D>0\end{array} $$ I could not find the correct range of values of a and b. Any help is appreciated!
Starting with a generic, factored cubic equation
$$\begin{align*}f(x) &= (x-r_0)(x-r_1)(x-r_2) \\ \\ &= x^2 -(r_0 + r_1 +r_2)x^2 + (r_0r_1 + r_0r_2+r_1r_2)x - r_0r_1r_2 \\ \end{align*}$$
In your specific case, you have $$\begin{align*}a &= (r_0 + r_1 +r_2) \\ b&= (r_0r_1 + r_0r_2+r_1r_2) \\ 2 &= r_0r_1r_2 \\ \end{align*}$$
Computing the product $ab$ $$\begin{align*}ab &= (r_0 + r_1 +r_2)(r_0r_1 + r_0r_2+r_1r_2) \\ \\ &= 3r_0r_1r_2 +r_0^2(r_1+r_2)+r_1^2(r_0+r_2)+ r_2^2(r_0+r_1)\\ \end{align*}$$
And now with a little hand-waving, I'll say that since the product of the 3 roots is fixed to a constant, all 3 roots are positive and implied to be real, and since the expression for $ab$ has symmetry with respect to the 3 roots, that the expression for $ab$ is minimized when the three roots are equal. (Similar to a given regular, right rectangular volume has minimal total edge length when all the dimensions are equal [a cube]; or a given rectangular area has minimum perimeter when all the sides are equal [a square].) $$\begin{align*}r &= r_0 = r_1 = r_2\\ \\ r^3 &= r_0r_1r_2 = 2\\ \\ \min{ab} &= 3 r^3 + 3r^2(2r)\\ \\ &= 9r^3\\ \\ \min{ab}&= 9(2) = 18 \end{align*}$$