Thanks for the answer!

That is indeed the system that I had to solve. The constraints on $a$ and $b$ are easy to handle. The constraint on $c$ is harder to handle. if Mathematica gives as output $\mbox{Root}[\ldots,1]$ and $\mbox{Root}[\ldots,2]$, then I assume the latter to be the greater (real) root?

When I only consider the first thee equations/inequalities of the systems then Mathematica also gives a constraint on $b$ in function of Roots objects. More precisely, one of the cases for example states $a^2 < b < \mbox{Root}[375 a^8 - 3600 a^6 x + 8088 a^4 x^2 - 6912 a^2 x^3 + 2048 x^4,2]$. I let Wolfram|Alpha solve the latter quartic and the two real solutions are: $0,147 a^2$ and $1.08148a^2$. Hence my example can also be written as: $$a^2<b<1,08148 a^2,$$ which is a fairly easy condition to work with.

Adding the last cubic inequality $4x^3 - 8 c x^2 + 5 a c x - c b <0$ to the system and solving it with Mathematica gives exactly the same constraints on $a$ and $b$ as above. Similarly, $c$ is also bounded by a Root object. For instance, $$\mbox{Root}[27b^3 +(27a^3 b - 162 a b^2)x + (-45a^4 + 225a^2 b + 72 b^2) x^2 + (-40 a^3 - 168 a b)x^3 + (48 a^2 + 16b) x^4]<c<0 $$ I also let Wolfram|Alpha solve the above quartic and the solution is horrendous. However, since I know the values of $a$ and $b$, I can easily compute the roots of the quartic with the help of Matlab for instance. Thus I think it's best to leave it like that?

Is there maybe a better way to obtain a condition on $c$ such that the whole system of equations/inequalities is satisfied? I mean, if I let Mathematica solve the cubic inequality then my condition on $c$ would be easier I guess, but that does not necessarily mean that the whole system is satisfied?

Thanks!