<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] 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.
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