Let
$$g(x)=ax^3+bx^2+cx+d$$
be a polynomial of degree $3$ with $a,b,c,d,e\in\mathbb{R}$ and $a>0$. Suppose that the cubic equation $g(x)=0$ has three distinct real roots, i.e. the discriminant $\Delta>0$.
Let $f(x)=\frac{a}{4}x^4+\frac{b}{3}x^3+\frac{c}{2}x^2+dx$. Can we express $M:=\min(f(r_1),f(r_2),f(r_3))$ in terms of the parameters $a,b,c,d$ without directly inserting $r_1$, $r_2$ and $r_3$ (given by solution formula) into $f(x)$ and compare them? In other words, we have explicit solution as shown, for instance in here. Can we give conditions on $a,b,c,d$ and thus determine which $r_i$ is the one that satisfies $M$?
Any reference, suggestion, idea, or comment is welcome. Thank you!
So, (lots of this is on the Wikipedia page linked in above comments), if we set $$x=t-\frac{b}{3a} ..[1]$$ then $g(x)$ becomes $G(t)=t^3+pt+q$ where $p=\frac{3ac-b^2}{3a^2}$ and $q=\frac{2b^3-9abc+27a^2d}{27a^3}$.
Provided $4p^3+27q^2<0$, it has three real roots given by setting $k=0, 1, 2$ in $$t_k=2\sqrt{\frac{-p}{3}}\cos\left(\frac13\cos^{-1}\left(\frac{3q}{2p}\sqrt{\frac{3}{-p}}\right)-\frac{2\pi k}{3}\right) ...[2]$$
Then the integral of G(t) becomes $$F(t)=\frac{t^4}{4}+\frac{pt^2}{2}+qt$$ which will only differ from f(x) by a constant. Assuming $a>0$, any deduction about which root gives the minimum value for F(t) will translate to the same conclusion for f(x).
So it remains to evaluate F(t) at each of the roots of G(t). As we are only concerned with the minimum value, we need only consider the lowest and highest root which are respectively given by $k=2$ and $k=0$ in [2]. (This can be established by recognising that if $\theta = \frac13\cos^{-1}\left(\frac{3q}{2p}\sqrt{\frac{3}{-p}}\right)$ then $0< \theta < \frac{\pi}{3}$).
By definition, at a root of G(t), $t^3 = -pt-q$ and so at the roots, we can show $$F(t_k)=\frac{t_k}{4}(pt_k+3q).$$ The two values of interest are $F(t_2)$ and $F(t_0)$. Once you have established which of these is lower, you can translate the relevant value of t into a value for x using [1] and evaluate f(x) there.
Given $0< \theta < \frac{\pi}{3}$, it is possible to show that $$F(t_2)>F(t_0)$$ $$\Leftarrow\Rightarrow \cos(\theta-\frac{4\pi}{3})+\cos(\theta)+\cos(3\theta)>0$$ $$\Leftarrow\Rightarrow \theta<\frac{\pi}{6}$$ $$\Leftarrow\Rightarrow q<0.$$
So, putting it all together, if $q<0$ ie if $$2b^3-9abc+27a^2d<0 $$ then the minimum value is given by $$f(t_0-\frac{b}{3a})$$ and if $$2b^3-9abc+27a^2d>0 $$ then the minimum value is given by $$f(t_2-\frac{b}{3a})$$