For all ordered triples $(p,q,r)$ define the polynomial
$$f_{p,q,r}(x)=x^3-px^2+qx-r$$
Let $a_{1},a_{2},a_{3},b_{1},b_{2},b_{3},c_{1},c_{2},c_{3}$ be (not necessarily distinct) positive reals such that the roots of $f_{a_{1},a_{2},a_{3}}(x)$ are $b_{1},b_{2},b_{3}$ and the roots of $f_{b_{1},b_{2},b_{3}}(x)$ are $c_{1},c_{2},c_{3}$. Determine the maximum possible value of
$$ \frac{9\sqrt[3]{b_{3}}}{b_{1}+3} + \frac{4+3b_{1}+2b_{2}+b_{3}}{a_{1}+1} $$
I used Vieta's formulas combined with calculus. I set this expression equal to $y$ and then cubed both sides. Then I tried to use the fact that since $y$ is real, the cubic in $y$ (generated by cubing both sides) will have three real roots. Now, I differentiated the equation w.r.t. $y$ (assuming everything else to be constant). I got a quadratic in $y$ and I then made its discriminant $>0$. Now I used Vieta's formulas. After that I'm stuck since I still have more than one variable left. Also, I'm not yet familiar with multivariable calculus. Any help will be greately appreciated. Thanks!
I believe that $$ \frac{9b_3^{1/3}}{b+1}+\frac{4+3b_1+2b_2+b_3}{a_1+1}\leq 4, $$ equality holds for $c_1=c_2=c_3=1$, $b_1=3$, $b_2=3$, $b_3=1$, $a_1=7$, $a_2=15$, $a_3=9$.
At first, we see that $c_1,c_2,c_3$ define all others: $$ b_1=c_1+c_2+c_3, $$ $$ b_2=c_1c_2+c_2c_3+c_3c_1, $$ $$ b_3=c_1c_2c_3, $$ $$ a_1=b_1+b_2+b_3=c_1+c_2+c_3+c_1c_2+c_2c_3+c_1c_2c_3. $$ Moreover, they all are positive for any positive $c_1,c_2,c_3$, and we have no restriction for $c_i$.
Put expressions via $c_i$ for $b_i$ and $a_i$ in formula we get: $$ \frac{9(c_1c_2c_3)^{1/3}}{c_1+c_2+c_3+3}+\frac{4+3(c_1+c_2+c_2)+2(c_1c_2+c_2c_3+c_3c_1)+c_1c_2c_3}{c_1c_2c_3+c_1c_2+c_2c_3+c_3c_1+c_1+c_2+c_3+1}. $$ I would like to denote $c_1+1=x$, $c_2+1=y$, $c_3+1=z$. One can check that we have: $$ \frac{9((x-1)(y-1)(z-1))^{1/3}}{x+y+z}+\frac{xyz+xy+yz+zx}{xyz}. $$