Example 6 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 $\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}}$.
(2002 National High School League Question)
Solution 1: Let $x_{1}=m-\frac{\lambda}{2}, x_{2}=m+\frac{\lambda}{2}, x_{3}>\frac{1}{2}\left(x_{2}+x_{1}\right)=m$.
Let $x_{3}=m+t(\lambda>0, t>0)$.
By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
a=-\left(x_{1}+x_{2}+x_{3}\right)=-(3 m+t), \\
b=x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=3 m^{2}+2 m t-\frac{\lambda^{2}}{4}, \\
c=-x_{1} x_{2} x_{3}=-m^{3}-m^{2} t+\frac{\lambda^{2}}{4} m+\frac{\lambda^{2}}{4} t,
\end{array}\right.
$$

Then $2 a^{3}-9 a b=a\left(2 a^{2}-9 b\right)=27 m^{3}+27 m^{2} t-\frac{27}{4} m \lambda^{2}-\frac{9}{4} \lambda^{2} t-2 t^{3}$,
$$
\begin{array}{l}
27 c=-27 m^{2}-27 m^{2} t+\frac{27}{4} m \lambda^{2}+\frac{27}{4} \lambda^{2} t \text {. } \\
\text { Thus, } s=\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}}=\frac{9 \lambda^{2} t-4 t^{3}}{2 \lambda^{3}} \text { (to achieve the maximum, } 9 \lambda^{2}-4 t^{2} \text { should be positive) } \\
=\frac{1}{4 \sqrt{2} \lambda^{3}} \sqrt{\left(9 \lambda^{2}-4 t^{2}\right)\left(9 \lambda^{2}-4 t^{2}\right) \times 8 t^{2}} \\
\leqslant \frac{1}{4 \sqrt{2} \lambda^{3}} \sqrt{\left[\frac{2\left(9 \lambda^{2}-4 t^{2}\right)+8 t^{2}}{3}\right]^{3}} \\
=\frac{3 \sqrt{3}}{2} \text {. } \\
\end{array}
$$

When and only when $9 \lambda^{2}-4 t^{2}=8 t^{2} \Rightarrow t=\frac{\sqrt{3}}{2} \lambda$ (at this point $9 \lambda^{2}-4 t^{2}>0$), if we set $m=0, \lambda^{2}=2$,
then $x_{1}=-1, x_{2}=1, x_{3}=\sqrt{3}$. It can be seen that when $a=-c=-\sqrt{3}, b=-1$, the maximum value $\frac{3 \sqrt{3}}{2}$ can be achieved.
$$
\begin{array}{l}
\text { Solution } 2 \text { Note that the numerator of the original expression } \\
2 a^{3}+27 c-9 a b=27\left[\left(-\frac{a}{3}\right)^{3}+a \cdot\left(-\frac{a}{3}\right)^{2}+b \cdot\left(-\frac{a}{3}\right)+c\right] \\
=27 f\left(-\frac{a}{3}\right)=27\left(-\frac{a}{3}-x_{1}\right)\left(-\frac{a}{3}-x_{2}\right)\left(-\frac{a}{3}-x_{3}\right) \\
=\left(-a-3 x_{1}\right)\left(-a-3 x_{2}\right)\left(-a-3 x_{3}\right) \text {. } \\
\text { Since }-a=x_{1}+x_{2}+x_{3} \text {, } \\
\text { the numerator }=\left(x_{2}+x_{3}-2 x_{1}\right)\left(x_{1}+x_{3}-2 x_{2}\right)\left(x_{1}+x_{2}-2 x_{3}\right) \\
\text { Let } x_{2}=x_{1}+2 m, x_{3}=x_{1}+m+n \text {, where } m>0, n>0 \text {, } \\
\text { then the numerator }=(3 m+n)(3 m-n) 2 n=2 n\left(9 m^{2}-n^{2}\right) \text {. } \\
\text { The original expression }=\frac{1}{4} \times\left[9 \times\left(\frac{n}{m}\right)-\left(\frac{n}{m}\right)^{3}\right] \text {. } \\
\text { Let } \frac{n}{m}=t \text {, then the original expression }=\frac{1}{4}\left(9 t-t^{3}\right), t>0 \text {. } \\
\text { Let } g(t)=9 t-t^{3} \text {, for } t_{1}>t_{2} \text { in } (0, \sqrt{3}] \text {, we have } \\
g\left(t_{1}\right)-g\left(t_{2}\right)=\left(t_{1}-t_{2}\right)\left[9-\left(t_{1}^{2}+t_{2}^{2}+t_{1} t_{2}\right)\right] \\
>\left(t_{1}-t_{2}\right)\left[9-\left((\sqrt{3})^{2}+(\sqrt{3})^{2}+\sqrt{3} \cdot \sqrt{3}\right)\right]=0 \text {. } \\
\end{array}
$$

Therefore, $g(t)$ is an increasing function in $(0, \sqrt{3}]$ and a decreasing function in $[\sqrt{3},+\infty)$.
Hence, at $t=\sqrt{3}$, $g(t)$ reaches its maximum value of