THE 2002 VIETNAMESE MATHEMATICAL OLYMPIAD

The demanded inequality can be written in the form: $$-6a(a^2-2b)\le -27c+10(a^2-2b{)}^{3/2}(1)$$ Let $\alpha ,\beta ,\gamma$ be three real roots of the polynomial $P(x)$. By the theorem of Viet, (1) is equivalent to $$6(\alpha +\beta +\gamma )({\alpha }^2+{\beta }^2+{\gamma }^2)\le 27\alpha \beta \gamma +10({\alpha }^2+{\beta }^2+{\gamma }^2{)}^{3/2}(2)$$ If ${\alpha }^2+{\beta }^2+{\gamma }^2=0$ then (2) is evident and in this case, (2) is an equality. If ${\alpha }^2+{\beta }^2+{\gamma }^2>0$, w. l. o. g. we can suppose that $$|\alpha |\le |\beta |\le |\gamma |(\ast )$$ $$\text{and}{\alpha }^2+{\beta }^2+{\gamma }^2=9(\ast \ast )$$ Then (2) is equivalent to $$2(\alpha +\beta +\gamma )-\alpha \beta \gamma \le 10(3)$$ We deduce from () and () that $$\begin{array}{rl}[2(\alpha +\beta +\gamma )-\alpha \beta \gamma {]}^2& =[2(\alpha +\beta )+\gamma (2-\alpha \beta ){]}^2\\ & \le [(\alpha +\beta {)}^2+{\gamma }^2][4+(2-\alpha \beta {)}^2]\\ & =(9+2\alpha \beta )[8-4\alpha \beta +(\alpha \beta {)}^2]\\ & =2(\alpha \beta {)}^3+(\alpha \beta {)}^2-20\alpha \beta +72\\ & =(\alpha \beta +2{)}^2(2\alpha \beta -7)+100\end{array}(4)$$ But from () and (), we see that ${\gamma }^2\ge 3$, therefore $2\alpha \beta \le {\alpha }^2+{\beta }^2=9-{\gamma }^2\le 6$ and (4) gives: $$[2(\alpha +\beta +\gamma )-\alpha \beta \gamma {]}^2\le 100,$$ so (3) is proved. (4) is an equality when and only when $$\{\begin{array}{l}|\alpha |\le |\beta |\le |\gamma |\\ {\alpha }^2+{\beta }^2+{\gamma }^2=9\\ (\alpha +\beta )/2=\gamma /2-\alpha \beta \\ \alpha \beta +2=0\\ 2(\alpha +\beta +\gamma )-\alpha \beta \gamma \ge 0\end{array}$$ therefore it occurs when and only when $\alpha =-1$ and $\beta =\gamma =2$. Thus (2) is an equality when and only when $(\alpha ,\beta ,\gamma )$ is a permutation of $(-t,2t,2t)$, where $t$ is a non negative real number, so (1) is an equality when and only when $$a=-3t,b=0\text{ and }c=4t^3,$$ where $t$ is a non negative real number.