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.

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 $\max (a c)^2 \lt 0.64$
