China Mathematical Competition (Extra Test)

Let $S=\frac{2a^3+27c-9ab}{{\lambda }^3}$, then $$\begin{gathered} S=\frac{27\left(\frac{2}{27}{a}^3-\frac{1}{3}ab+c\right)}{{\lambda }^3}=\frac{27f\left(-\frac{1}{3}a\right)}{{\lambda }^3} \\ =\frac{27\left(-\frac{1}{3}a-x_1\right)\left(-\frac{1}{3}a-x_2\right)\left(-\frac{1}{3}a-x_3\right)}{(x_2-x_1{)}^3} \\ =\frac{-27\left(x_1+\frac{1}{3}a\right)\left(x_2+\frac{1}{3}a\right)\left(x_3+\frac{1}{3}a\right)}{(x_2-x_1{)}^3}. \end{gathered}$$

Write $u_i=x_i+\frac{a}{3}$ ($i=1,2,3$), then $u_2-u_1=x_2-x_1=\lambda$, $u_3>\frac{1}{2}(u_1+u_2)$, and $u_1+u_2+u_3=x_1+x_2+x_3+a=0$. So, $u_1$, $u_2$ and $u_3$ satisfy the corresponding conditions too, and $$S=\frac{-27u_1{u}_2{u}_3}{(u_2-u_1{)}^3}.$$ By $u_3=-(u_1+u_2)>\frac{u_1+u_2}{2}$, we get $u_1+u_2<0$. Consequently, at least one of $u_1$ and $u_2$ should be less than $0$. We may assume $u_1<0$.

If $u_2<0$, then $S<0$.

If $u_2>0$, we suppose further $$v_1=\frac{-u_1}{u_2-u_1},v=v_2=\frac{u_2}{u_2-u_1}.$$ Then $v_1+v_2=1$, $v_1$ and $v_2$ are both greater than $0$ and $v_1-v_2=\frac{u_3}{u_2-u_1}>0$. So it follows that $$\begin{array}{rl}S& =27v_1{v}_2(v_1-v_2)=27v(1-v)(1-2v)\\ & =27\sqrt{(v-v^2{)}^2(1-2v{)}^2}\\ & =27\sqrt{2(v-v^2)(v-v^2)\left(\frac{1}{2}-2v+2v^2\right)}\\ & \le 27\sqrt{2\times {\left(\frac{1}{6}\right)}^3}\\ & =\frac{3}{2}\sqrt{3}.\end{array}$$ The equality holds when $v=\frac{1}{2}\left(1-\frac{\sqrt{3}}{3}\right)$. The corresponding cubic equation is $x^3-\frac{1}{2}x+\frac{\sqrt{3}}{18}=0$ and $\lambda =1$ ($x_1=-\frac{1}{2}(1+\frac{\sqrt{3}}{3})$, $x_2=\frac{1}{2}(1-\frac{\sqrt{3}}{3})$ and $x_3=\frac{\sqrt{3}}{3}$).

Consequently, the maximal value is $\frac{3}{2}\sqrt{3}$.