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.
 A: That $b$ is a root implies $-c = b^3+2ab^2+b^2+2a^2b$. Then $x^3+2ax^2+(2a^2+b)x+c = (x-b)(x^2+(2a+b)x+(2a^2+b+2ab+b^2)$ is valid. Since both roots of the quadratic are real, the discriminant must be non-negative: $(2a+b)^2-4(2a^2+b+2ab+b^2) \ge 0$, which is equivalent to $a \in [\frac{-b-\sqrt{-2b^2-4b}}{2},\frac{-b+\sqrt{-2b^2-4b}}{2}]$. We therefore must have $b \le 0$. $\bf\text{For ease, replace $b$ by $-b$}$. 
Our problem comes down to maximizing $$(ab^3-2a^2b^2-ab^2+2a^3b)^2$$ given that $$a \in [\frac{b-\sqrt{4b-2b^2}}{2},\frac{b+\sqrt{4b-2b^2}}{2}].$$ Note that we in particular must have $b \in [0,2]$. Let $f(a) = ab^3-2a^2b^2-ab^2+2a^3b$. Then $f'(a) \ge 0$ if and only if $6a^2-4ba+b^2-b \ge 0$, which has roots $\frac{2b\pm\sqrt{6b-2b^2}}{6}$. Now, for any $b \in [0,2]$, $\frac{2b-\sqrt{6b-2b^2}}{6} < \frac{2b+\sqrt{6b-2b^2}}{6} < \frac{b+\sqrt{4b-2b^2}}{2}$, and it holds that $\frac{b-\sqrt{4b-2b^2}}{2} < \frac{2b-\sqrt{6b-2b^2}}{6}$ if and only if $ b \in [0,\frac{50}{33})$. Since we want to maximize $f(a)^2$, it suffices to check extremal $a$ and $a$ for which $f'(a) = 0$, since $(f^2)' = 2ff'$ and $f(a)=0$ implies $(ac)^2 = 0 \le 3$. That is, we just have to show $f(a)^2 \le \sqrt{3}$ for $a = \frac{b+\sqrt{4b-2b^2}}{2},\frac{2b+\sqrt{6b-2b^2}}{6},\frac{2b-\sqrt{6b-2b^2}}{6}$, and for $a = \frac{b-\sqrt{4b-2b^2}}{2}$ when $b \in (\frac{50}{33},2]$. 
And this is easily doable. For example, at $a = \frac{b+\sqrt{4b-2b^2}}{2}$, $f(a) = \frac{1}{4}(2-b)b^2(b+\sqrt{2b(2-b)})$. Since $(2-b)b \le 1$, $f(a) \le \frac{1}{4}b(b+\sqrt{2})$, which is $\le \sqrt{3}$. The other values of $a$ can also be handled easily.
A: Hint.
If $b$ is a root then
$$
(x-b)(x-r_1)(x-r_2) = x^3+2ax²+(2a^2+b)x+c
$$
and comparing polynomials we get
$$
\left\{
\begin{array}{rcl}
b r_1 r_2 + c & = & 0\\
r_1 r_2 +b(r_1+r_2) & = & b(1-r_1)\\
r_1+r_2 + 2a+b & = & 0
\end{array}
\right.
$$
and after solving for $a,b,c$ we have
$$
a = \frac 12\left(1-r_1+r_2\pm\sqrt{1-r_1^2-r_2^2}\right)\\
b = -1\mp\sqrt{1-r_1^2-r_2^2}\\
c = r_1r_2\left(1\pm\sqrt{1-r_1^2-r_2^2}\right)
$$
so $a(r_1,r_2)c(r_1,r_2)$ have a surface over the circle $r_1^2+r_2^2\le 1$ represented respectively as follows:


The next step is the minima/maxima determination.
NOTE
In polar coordinates we have respectively
$$
\cases{
(a c)_1 = \frac{1}{4} \rho ^2 \sin (2 \theta ) \left(\left(\sqrt{1-\rho ^2}+1\right) (2-\rho  (\sin (\theta )+\cos (\theta )))-\rho
   ^2\right)\\
\\
(a c)_2 = \frac{1}{4} \rho ^2 \sin (2 \theta ) \left(\left(\sqrt{1-\rho ^2}-1\right) \rho  (\sin (\theta )+\cos (\theta )-2)-\rho ^2\right)
}
$$
now assuming $\max \frac{1}{4} \rho ^2 \sin (2 \theta )=\frac 14$ we have
$$
\max (a c)^2 \le 1.45
$$
with much algebraic effort we can conclude that $\max (a c)^2 \lt 0.64$
A: ►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
$$r^2sb+s^2rb+b^2rs\le 2\sqrt3$$
