Let $x_1$, $x_2$ and $x_3$ be the real roots of the equation $x^3 - a x^2 + b x - a = 0$. By Vieta's Formula, we have
$x_1 + x_2 + x_3 = a$, $x_1 x_2 + x_2 x_3 + x_1 x_3 = b$, $x_1 x_2 x_3 = a$.

By $(x_1 + x_2 + x_3)^2 \ge 3(x_1 x_2 + x_2 x_3 + x_1 x_3)$, we have $a^2 \ge 3b$, and by $a = x_1 + x_2 + x_3 \ge 3 \sqrt[3]{x_1 x_2 x_3} = 3 \sqrt[3]{a}$, we have $a \ge 3 \sqrt{3}$.

Thus,
$$\begin{aligned} \frac{2a^3 - 3ab + 3a}{b+1} &= \frac{a(a^2 - 3b) + a^3 + 3a}{b+1} \\ &\ge \frac{a^3 + 3a}{b+1} \ge \frac{a^3 + 3a}{\frac{a^2}{3} + 1} \\ &= 3a \ge 9\sqrt{3}. \end{aligned}$$
If $a = 3\sqrt{3}$, $b = 9$, then the equality holds when each root is equal to $\sqrt{3}$.

Summing up, the answer is $9\sqrt{3}$.