sqing wrote: Let $x_1,x_2,x_3$ be three positive real roots of the equation $x^3+ax^2+bx+c=0$ $(a,b,c\in R)$ and $x_1+x_2+x_3\leq 1. $ Prove that $$a^3(1+a+b)-9c(3+3a+a^2)\leq 0$$ Let $x_1+x_2+x_3=3u$, $x_1x_2+x_1x_3+x_2x_3=3v^2$ and $x_1x_2x_3=w^3$. Thus, $0<u\leq\frac{1}{3},$ $a=-3u$, $b=3v^2$, $c=-w^3$ and we need to prove that: $$u^3(3v^2-3u+1)\geq w^3(3u^2-3u+1)$$or $$(u^3-w^3)(3u^2-3u+1)\geq3u^3(u^2-v^2)$$and since by AM-GM $$u^3-w^3\geq u(u^2-v^2),$$it's enough to prove that $$3u^2-3u+1\geq3u^2$$or $$u\leq\frac{1}{3}.$$

This is G11 P1.

Note that $a^3\le 27c$(Use Vieta and note the minus sign in Vieta changes the direction) and $ab\le 9c$ (also Vieta) The other condition gives $1+a\ge 0$ So $$a^3(1+a+b)-9c(3+3a+a^2)=(1+a)(a^3-27c)+a^2(ab-9c)\le 0$$