Maximize product $(a^3-a^2+2)(b^3-b^2+2)(2c^3+5c^2+9)$ Consider three real variables such that $a^2+5b^2+5c^2 = 21$. Maximize the product:
$$P = (a^3-a^2+2)(b^3-b^2+2)(2c^3+5c^2+9)$$
My attempt (ideas): I suppose the first step is to give an argument that the maximum is reached when the variables are positive (or at least that's what I suspect). $a$ lies in $[-\sqrt{21}, \sqrt{21}]$, while $b$ and $c$ lie in $\left[-\sqrt{\dfrac{21}{5}}, \sqrt{\dfrac{21}{5}}\right]$. So I guess studying how the functions $f(x) = x^3-x^2+2$, $g(x)=2x^3+5x^2+9$ behave on those ranges should help, but I don't how to give a solid argument for this.
After this, I guess I should factor out:
$$P = (a+1)(b+1)(c+3)(a^2-2a+2)(b^2-2b+2)(2c^2-c+3)$$
and find some tricky way to use AM-GM, but it's difficult to find the correct way, without knowing the point for which the maximum is attained.
 A: Incomplete solution:  (Updated 2020-01-25)
Step 1 (incomplete): Establish that the maximum of $P$ occurs when all three factors of $P$ are non-negative.
First, let's observe that from the condition $a$ lies in $[-\sqrt{21},\sqrt{21}]$, while $b$ and $c$ lie in $\left[-\sqrt{\dfrac{21}{5}}, \sqrt{\dfrac{21}{5}}\right]$, so let's define the functions:
$$f_1 : \left[-\sqrt{21}, \sqrt{21}\right] \to \mathbb{R},\ f_1(x) = x^3-x^2+2$$
$$f_2 : \left[-\sqrt{\frac{21}{5}}, \sqrt{\frac{21}{5}}\right]\to \mathbb{R},\ f_2(x) = x^3-x^2+2$$
$$g : \left[-\sqrt{\frac{21}{5}}, \sqrt{\frac{21}{5}}\right]\to \mathbb{R},\ g(x) = 2x^3+5x^2+9 $$
We have 
$$P = f_1(a) \cdot f_2(b)\cdot g(c)$$
The maximum of $P$ is a positive value, so we need either all three factors of $P$ to be positive or two of them negative and the third one positive. Studying $g(x)$, we can see that this function is positive over it's entire definition domain. So $f_1(a)$ and $f_2(b)$ must both be either positive or negative.
Assuming that $f_1(a)$ and $f_2(b)$ are both negative, we should prove that the maximum reachable value is lower than if $f_1(a)$ and $f_2(b)$ are both positive.
Step 2.1 (thanks to @Maximilian Janisch comment): Checking $(a,b,c) = (4,0,1)$, we can see that $P$ equals $1600$. We will prove that $P \leq 1600$.
Step 2.2: Notice that using AM-GM:
$$x^3-x^2+2=(x+1)(x^2-2x+2)=\frac{1}{2}(2x+2)(x^2-2x+2)\leq \frac{1}{8}(x^2+4)^2$$
This usage of AM-GM is satisfactory because it conserves both equality cases for $a$ and $b$ ($4$ and $0$). Similarly:
$$2x^3+5x^2+9=(x+3)(2x^2-x+3) \leq \frac{1}{4}(2x^2+6)^2=(x^2+3)^2$$
Therefore, using these observations and AM-GM again:
$$
\begin{aligned}P &\leq \frac{1}{64} \left[(a^2+4)(b^2+4)(c^2+3)\right]^2 \\
    &= \frac{1}{64} \left[\frac{1}{25}(a^2+4)\cdot 5(b^2+4)\cdot 5(c^2+3) \right]^2 \\
    &\leq \frac{1}{64} \left[\frac{1}{25}\cdot \frac{1}{27} (a^2+5b^2+5c^2+39)^3 \right]^2\\
    &= 1600\end{aligned} $$
