Vieta equality and inequality

$7x^3-ax^2+bx-12=0$ Using Vieta's Formula , $\;\;\;\;\;\;\;\;\;\; r_1+r_2+r_3=\frac{a}{7}$ $r_1r_2+r_2r_3+r_3r_1=\frac{b}{7}$ $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; r_1r_2r_3=\frac{12}{7}$ Applying the AM-GM inequality , $3=\frac{7r_1}{2}+\frac{r_2}{3}+\frac{r_3}{2} \ge 3\sqrt[3]{\frac{7r_1}{2}\cdot \frac{r_2}{3}\cdot \frac{r_3}{2}}=3$ So, by the equality condition of AM-GM, $\frac{7r_1}{2}=\frac{r_2}{3}=\frac{r_3}{2}=1 \implies r_1=\frac{2}{7}\;, \;\; r_2=3\;, \;\;r_3=2$ $\implies a=7( r_1+r_2+r_3)=37$ $\;\;\;\;\;\;\;\;\;\;\implies b=7(r_1r_2+r_2r_3+r_3r_1)=52$ Therefore, $a=37, b=52\implies \boxed{a+b=89}$