Given the system of equations/inequalities: $$\left \{ \begin{array}{llll} 16x^4-40ax^3+(15a^2+24b)x^2-18abx+3b^2 = 0 \\ 5ax-4x^2-b>0 \\ 15ax-20x^2-3b<0, \end{array}\right.$$ where $x<0$, and $a<0,c<0$. In fact, I'm not particularly interested in the solution $x$ itself but more in the constraints on $(a,b)$ such that the system admits a solution. With the help of Wolfram|Alpha I came to the conclusion that the system admits a real solution $x$ whenever: $$ a \neq 0, 0<b<\frac{a^2}{32}(27+\sqrt{111+46 \sqrt{6}}-3\sqrt{6}) = 1.08148 a^2 $$
Now, proceeding, it appears that $x$ also satisfies $$4x^3 - 8cx^2 + 5 ca x - cb <0 \qquad (*),$$ where $c<0$ is an additional parameter. Therefore to find $x$ I have to extend the original system: $$\left \{ \begin{array}{llll} 16x^4-40ax^3+(15a^2+24b)x^2-18abx+3b^2 = 0 \\ 5ax-4x^2-b>0 \\ 15ax-20x^2-3b<0 \\ 4x^3-8cx^2+5cax-cb<0 \end{array} (**)\right.$$
The objective is to extend my constraints on $(a,b)$ from above with an additional constraint on $c$ such that the new system yields a real solution $x$. I tried to run it in Wolfram|Alpha. I'm not sure why, but sometimes it is able to compute a solution and sometimes it does not (perhaps it's due to the server). However, the new constraints on $a$ and $b$ are fairly simple to handle, but the constraint on $c$ is horrendous.
Therefore, my question. Is there better way to impose the new constraints on $(a,b,c)$. I mean, I have fairly easy conditions on $a$ and $b$, so it's a little bit unfortunate that the new cubic inequality ruins it. I can solve the cubic inequality separately from the others which gives me an interval for $x$ and easy constraints on $(a,b,c)$. But then, I did not find a solution that satisfies all the equations and inequalities. Or is there a better way?
Ps: Is there perhaps someone with Mathematica who could solve the new system $(**)$?
Thanks in advance! Cheers.
I managed to let Mathematica solve the system. The conditions on $a$ and $b$ do not change by adding the cubic. However, the condition on $c$ is very messy.
Solving the cubic inequality separately from the others, gives me a much easier constraint on $c$ I guess, but I think that it does not necessarily imply that I can use is as a constraint for the whole system?
So is there a way to guarantee that solving the cubic inequality separately also implies a solution for the whole system?