<s>[INST] The polynomial $x^3 + 2ax^2 + (2a^2 + b)x + c$ has three (real and not necessairily distinct) roots and $b$ is one of them. Prove that $(ac)^2 \le 3$. 
The polynomial $$\large x^3 + 2ax^2 + (2a^2 + b)x + c$$ has three (real and not necessairily distinct) roots and $b$ is one of them. Prove that $(ac)^2 \le 3$.

I'm uncertain of how to prove this.
If $b$ is a root of $x^3 + 2ax^2 + (2a^2 + b)x + c$ then $b$ is also a root of $$(2a + b)x^2 + 2a^2x + (b^2 + c) = 0$$
which means the above polynomial has at least one root $\implies (a^2)^2 - (2a + b)(b^2 + c) \ge 0$
$\iff a^4 - 2ab^2 - 2ca - b^3 - bc \ge 0$
And $b$ is also a root of $x^3 + 2ax^2 + (2a^2 + b)x + c$ then $b$ is also a root of $$2ax^2 + (2a^2 + b)x + (b^3 + c) = 0$$
which means the above polynomial has at least one root $\implies (2a^2 + b)^2 - 4 \cdot 2a(b^3 + c) \ge 0$
$\iff 4a^4 + 4a^2b - 8ab^3 - 8ca + b^2 \ge 0$
But that's all I got.
 [/INST] ►In the case of three equal roots, it is proved without difficulty that there are only two possible cases $f(x)=x^3$ and $f(x)=(x+\frac23)^3$.
►For three roots $b,r,r$ we have 
$$f(x)=x^3-(b+2r)x^2+(2br+r^2)x-br^2\\f(x)=x^3+2ax^2+(2a^2+b)x+c$$ then$$\begin{cases}b+2r=-2a\\2br+r^2=2a^2+b\\-br^2=c\end{cases}\Rightarrow b^2+2b+2r^2=0\Rightarrow b\lt0$$ 
All the possible $f(x)$ in this second condition with the roots $b,r,r$ are such that the points $(b,r)$ are in the ellipse centered at $(-1,0)$ and having axes $1$ and $\dfrac{1}{\sqrt2}$.
On the other hand 
$$(ac)^2=\left(\frac{2r+b}{-2}\right)^2(-br^2)^2=\frac{b^2r^4(2r+b)^2}{4}$$
so, equivalently, we can prove $$b^2r^4(2r+b)^2\le12\qquad(*)$$
Taking the parametrics of the ellipse above we have $(b,r)=\left(\cos(t)-1,\dfrac{\sin(t)}{\sqrt2}\right)$ and the inequality (*) becomes
$$g(t)=(\cos(t)-1)^2\sin^4(t)(\sqrt2\sin(t)+\cos(t)-1)^2\le{24}$$
It is clear that $g(t)$ has an absolute maximum less than $24$ (exactly this absolute maximum is equal to $10.074$ and is taken at point $t=4.325$).
►For three distinct roots $b,r,s$ we have similarly the relation
$$b^2+r^2+s^2+2b=0$$ and we have to prove equivalently that
$$r^2sb+s^2rb+b^2rs\le 2\sqrt3$$
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