1. Real numbers $a, b$ make the equation
$$
x^{3}-a x^{2}+b x-a=0
$$

have three positive real roots. Find the minimum value of $\frac{2 a^{3}-3 a b+3 a}{b+1}$.
1. Let the three positive real roots of equation (1) be $x_{1}, x_{2}, x_{3}$. Then, by the relationship between roots and coefficients, we have
$$
\begin{array}{l}
x_{1}+x_{2}+x_{3}=a, \\
x_{1} x_{2}+x_{2} x_{3}+x_{1} x_{3}=b, \\
x_{1} x_{2} x_{3}=a .
\end{array}
$$

Thus, $a>0, b>0$.
$$
\begin{array}{l}
\text { By }\left(x_{1}+x_{2}+x_{3}\right)^{2} \geqslant 3\left(x_{1} x_{2}+x_{2} x_{3}+x_{1} x_{3}\right) \\
\Rightarrow a^{2} \geqslant 3 b . \\
\text { Also, } a=x_{1}+x_{2}+x_{3} \geqslant 3 \sqrt[3]{x_{1} x_{2} x_{3}}=3 \sqrt[3]{a} \\
\Rightarrow a \geqslant 3 \sqrt{3} . \\
\text { Then } \frac{2 a^{3}-3 a b+3 a}{b+1}=\frac{a\left(a^{2}-3 b\right)+a^{3}+3 a}{b+1} \\
\geqslant \frac{a^{3}+3 a}{b+1} \geqslant \frac{a^{3}+3 a}{\frac{a^{2}}{3}+1}=3 a \geqslant 9 \sqrt{3} .
\end{array}
$$

When $a=3 \sqrt{3}, b=9$, i.e., the three roots of equation (1) are all $\sqrt{3}$, the equality holds.
In summary, the minimum value sought is $9 \sqrt{3}$.