China Southeastern Mathematical Olympiad

Let $x_1$, $x_2$ and $x_3$ be the real roots of the equation $x^3-a{x}^2+bx-a=0$. 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, $$\begin{array}{rl}\frac{2a^3-3ab+3a}{b+1}& =\frac{a(a^2-3b)+a^3+3a}{b+1}\\ & \ge \frac{a^3+3a}{b+1}\ge \frac{a^3+3a}{\frac{a^2}{3}+1}\\ & =3a\ge 9\sqrt{3}.\end{array}$$ If $a=3\sqrt{3}$, $b=9$, then the equality holds when each root is equal to $\sqrt{3}$.

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