11. (15 points) Let real numbers $a, b$ be such that the equation $a x^{3}-x^{2} + b x-1=0$ has three positive real roots. For all real numbers $a, b$ that satisfy the condition, find the minimum value of $P=\frac{5 a_{2}^{2}-3 a b+2}{a^{2}(b-a)}$.


11. Solution: On the one hand, when $a=\frac{\sqrt{3}}{9}, b=\sqrt{3}$, the equation $\frac{\sqrt{3}}{9} x^{3} - x^{2} + \sqrt{3} x - 1 = 0 \Leftrightarrow (x - \sqrt{3})^{3} = 0$, so this equation has three equal positive real roots $\sqrt{3}$. At this time, $P = 12 \sqrt{3}$. Below, we prove that $12 \sqrt{3}$ is the minimum value sought.

Let the three positive real roots of the equation $a x^{3} - x^{2} + b x - 1 = 0$ be $\alpha, \beta, \gamma$, respectively. Then, by the relationship between roots and coefficients, we have: $\alpha + \beta + \gamma = \frac{1}{a}, \alpha \beta + \beta \gamma + \gamma \alpha = \frac{b}{a}, \alpha \beta \gamma = \frac{1}{a}$. Therefore, $a > 0, b > 0$.
From $(\alpha + \beta + \gamma)^{2} \geqslant 3(\alpha \beta + \beta \gamma + \gamma \alpha)$, we know that $\frac{1}{a^{2}} \geqslant \frac{3 b}{a}$, which gives: $b \leqslant \frac{1}{3 a}$.
Also, from $\alpha \beta + \beta \gamma + \gamma \alpha \geqslant 3 \sqrt[3]{\alpha^{2} \beta^{2} \gamma^{2}}$, we know that $\frac{b}{a} \geqslant 3 \sqrt[3]{\frac{1}{a^{2}}}$, which gives: $b \geqslant 3 \sqrt[3]{a}$. Therefore, we have $3 \sqrt[3]{a} \leqslant b \leqslant \frac{1}{3 a}$, so $3 \sqrt[3]{a} \leqslant \frac{1}{3 a}$, solving this gives $a \leqslant \frac{\sqrt{3}}{9}$, hence $a = \sqrt[3]{a^{2}} \sqrt[3]{a}0$, so $00$, so we get $P \geqslant \frac{15 a^{2} + 3}{a - 3 a^{3}} \geqslant 12 \sqrt{3}$.
In summary, the minimum value of $P$ is $12 \sqrt{3}$.