China Girls' Mathematical Olympiad

Proof We denote by $x_1,x_2,x_3$ the three positive roots of the polynomial $\phi (x)=a{x}^3+b{x}^2+cx+d$. By Vieta's theorem, we have $$\begin{array}{rl}{x}_1+x_2+x_3& =-\frac{b}{a},\\ x_1{x}_2+x_2{x}_3+x_3{x}_1& =\frac{c}{a},\\ x_1{x}_2{x}_3& =-\frac{d}{a}.\end{array}$$ --- As $\phi (0)<0$, we get $d<0$, and hence $a>0$. Dividing both sides of ① by $a^3$, we get an equivalent form of ①: $$\begin{array}{rl}& 7\left(-\frac{b}{a}\right)\frac{c}{a}\le 2{\left(-\frac{b}{a}\right)}^3+9\left(-\frac{d}{a}\right)\\ \iff & 7(x_1+x_2+x_3)(x_1{x}_2+x_2{x}_3+x_3{x}_1)\\ & \le 2(x_1+x_2+x_3{)}^3+9x_1{x}_2{x}_3\\ \iff & x_1^2{x}_2+x_1^2{x}_3+x_2^2{x}_1+x_2^2{x}_3+x_3^2{x}_1+x_3^2{x}_2\\ & \le 2(x_1^3+x_2^3+x_3^3).\stackrel{◯}{2}\end{array}$$ Because $x_1,x_2,x_3$ are all greater than 0, $(x_1-x_2)(x_1^2-x_2^2)\ge 0$. That is to say, $$x_1^2{x}_2+x_2^2{x}_1\le x_1^3+x_2^3.$$ By the same argument, $x_2^2{x}_3+x_3^2{x}_2\le x_2^3+x_3^3$, $x_3^2{x}_1+x_1^2{x}_3\le x_3^3+x_1^3$. Summing up these three inequalities we get ②, and the equality holds iff $x_1=x_2=x_3$.