Let the roots be $ r_1,r_2,r_3$. Then, since $ \frac{-d}{a}=r_1r_2r_3>0$ and $ d<0$, we have $ a>0$. Thus, we can divide the inequality by $ a^3$ to obtain an equivalent inequality: $ 2\frac{b^3}{a^3}+9\frac{d}{a}-7\frac{b}{a}\frac{c}{a}\le 0$ $ \Leftrightarrow -2(r_1+r_2+r_3)^3-9(r_1r_2r_3)+7(r_1+r_2+r_3)(r_1r_2+r_2r_3+r_3r_1)\le0$ Since this inequality is homogeneous, we can assume $ r_1+r_2+r_3=1$ to obtain $ -2-9r_1r_2r_3+7(r_1r_2+r_2r_3+r_3r_1)\le0$ $ \Leftrightarrow -\frac{2}{9}-\frac{7^3}{9^3}+\frac{49}{81}(r_1+r_2+r_3)-r_1r_2r_3+\frac{7}{9}(r_1r_2+r_2r_3+r_3r_1)\le -\frac{7^3}{9^3}+\frac{49}{81}(r_1+r_2+r_3)$ $ \Leftrightarrow \prod\left(\frac{7}{9}-r_i\right)\le \frac{64}{729}$ Now, if the right hand side is negative, we're done. If not, either all or one factor is positive. Obviously, there can't be two terms negative since then there would be two roots greater than $ \frac{7}{9}$ and so the sum of the roots is greater than one. If all factors are positive then by AM-GM, $ \prod\left(\frac{7}{9}-r_i\right)\le \left(\frac{1}{3}\sum\frac{7}{9}-r_i\right)^3=\frac{64}{729}$ as desired.

cosinator wrote: $ \Leftrightarrow - 2(r_1 + r_2 + r_3)^3 - 9(r{}_1r{}_2r{}_3) + 7(r_1 + r_2 + r_3)(r{}_1r{}_2 + r{}_2r{}_3 + r{}_3r{}_1)\le0$ You can just expand from here and cancel stuff. Substitute $ - \frac {b}{a} = x_1 + x_2 + x_3$, $ \frac {c}{a} = x{}_1x{}_2 + x{}_2{}_3 + x{}_3x{}_1$, $ - \frac {d}{a} = x{}_1x{}_2x{}_3$, and the inequality is equivalent to $ x_1^2x_2 + x{}_1x{}_2^2 + x_2^2x_3 + x{}_2x{}_3^2 + x_3^2x_1 + x{}_3x{}_1^2\leq 2(x_1^3 + x_2^3 + x_3^3)$, true by Schur.

cosinator wrote: for $ r_1+r_2+r_3 = 1$: $ - 2 - 9r_1r_2r_3 + 7(r_1r_2 + r_2r_3 + r_3r_1)\le0$ This is British MO 1999, Round 2, Question 3.

minsoens wrote: You can just expand from here and cancel stuff. Substitute $ - \frac {b}{a} = x_1 + x_2 + x_3$, $ \frac {c}{a} = x{}_1x{}_2 + x{}_2{}_3 + x{}_3x{}_1$, $ - \frac {d}{a} = x{}_1x{}_2x{}_3$, and the inequality is equivalent to $ x_1^2x_2 + x{}_1x{}_2^2 + x_2^2x_3 + x{}_2x{}_3^2 + x_3^2x_1 + x{}_3x{}_1^2\leq 2(x_1^3 + x_2^3 + x_3^3)$, true by Schur. This is also trivial by Muirhead (and equivalently true by AM-GM). $ \sum_\text{sym}r_1^2r_2\le\sum_\text{sym}\frac{r_1^3+r_2^3+r_2^3}{3}$

how can we write r1+r2+r3=1?????

We can always normalize homogeneous inequalities.