11. (2002 National Competition Question) Real numbers $a, b, c$ and a positive number $\lambda$ make $f(x)=x^{3}+a x^{2}+b x+c$ have three real roots $x_{1}, x_{2}, x_{3}$, and satisfy
(1) $x_{2}-x_{1}=\lambda$; (2) $x_{3}>\frac{1}{2}\left(x_{1}+x_{2}\right)$.

Find the maximum value of $\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}}$.
11. Solution: Let $x_{3}=\frac{1}{2}\left(x_{1}+x_{2}\right)+t(t>0)$.

By Vieta's formulas, we have $\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}}=\frac{-2\left(x_{1}+x_{2}+x_{3}\right)^{3}+27 x_{1} x_{2} x_{3}+9\left(x_{1}+x_{2}+x_{3}\right)\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}\right)}{\lambda^{3}}$
$$=\frac{\left(x_{1}+x_{2}-2 x_{3}\right)\left(x_{2}+x_{3}-2 x_{1}\right)\left(x_{3}+x_{1}-2 x_{2}\right)}{\lambda^{3}}$$
(When $x_{3}$ is replaced by $\frac{x_{1}+x_{2}}{2}$ in the numerators of the above two expressions, both become 0. By symmetry, the above equation holds)
$$\begin{array}{l}
=\frac{t(2 t+3 \lambda)(3 \lambda-2 t)}{2 \lambda^{3}} \text{ (substituting (1) into the equation) } \\
\leqslant \frac{\sqrt{t^{2}\left(9 \lambda^{2}-4 t^{2}\right)^{2}}}{2 \lambda^{3}}=\frac{\sqrt{8 t^{2}\left(9 \lambda^{2}-4 t^{2}\right)\left(9 \lambda^{2}-4 t^{2}\right)}}{4 \sqrt{2} \lambda^{3}} \\
\leqslant \frac{1}{4 \sqrt{2} \lambda^{3}} \sqrt{\left(\frac{18 \lambda^{2}}{3}\right)^{3}}=\frac{3 \sqrt{3}}{2} . \text{ The equality holds when } a=-c=-\sqrt{3}, b=-1.
\end{array}$$