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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.

LHF
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    The maximum seems to occur at $a=4, b=0, c=1$ and $P(4,0,1)=1600$. A proof of this might be pretty hard – Maximilian Janisch Jan 24 '20 at 18:50
  • @MaximilianJanisch, that helps a lot. I've found how to apply AM-GM (should I edit my attempt in the question or post it as an incomplete answer?). All that I need now is an argument for the non-negativity of $a$, $b$ and $c$. – LHF Jan 24 '20 at 19:05
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    I think posting it as an (incomplete) answer is appropriate. For the negative arguments: Unfortunately, even $P(-a,-b,-c)\le P(a,b,c)$ is not always true. All that I can say is that it seems that if at least one of the three numbers is negative, then the maximum is attained near $a=\frac92,b=0,c=-\sqrt{\frac{3}{20}}$ with $$P\left(\frac92,0,-\sqrt{\frac{3}{20}}\right)=\frac{1749}{400} \left(325-\sqrt{15}\right)\approx1404$$ – Maximilian Janisch Jan 24 '20 at 19:14
  • You could always parametrise the constraint as $a = \sqrt{21} \sin(\alpha)$, $b = \sqrt{\frac{21}{5}} \sin(\beta)\cos(\alpha)$ and $c = \sqrt{\frac{21}{5}} \cos(\beta)\cos(\alpha)$. Rewrite $P$ in terms of $\alpha$ and $\beta$, and it becomes an unconstrained maximisation problem. Lot of calculation though. – Lukas Rollier Jan 24 '20 at 19:21

2 Answers2

1

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:

$$ \begin{aligned}P &\leq \frac{1}{64} \left[(a^2+4)(b^2+4)(c^2+3)\right]^2 \\ &= \frac{1}{64} \left[\frac{1}{25}(a^2+4)\cdot 5(b^2+4)\cdot 5(c^2+3) \right]^2 \\ &\leq \frac{1}{64} \left[\frac{1}{25}\cdot \frac{1}{27} (a^2+5b^2+5c^2+39)^3 \right]^2\\ &= 1600\end{aligned} $$

LHF
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  • Note that $ab\le \frac{(a+b)^2}4$ holds for all reals $a,b$ so I don’t think your proof uses $a,b,c>0$ – Maximilian Janisch Jan 24 '20 at 19:49
  • @Maximilian Janisch, the observations I wrote hold true for any real $x$. However, there is a problem when I multiply $a^3-a^2+2 \leq \frac{1}{8}(a^2+4)^2$ with $b^3-b^2+2 \leq \frac{1}{8}(b^2+4)^2$. The resulting inequality would be correct only if $x^3-x^2+2 \geq -\frac{1}{8}(x^2+4)^2$ would also be true, however this is not the case over the entire range of $a,b$. Similar problem with the other observation for $c$. – LHF Jan 24 '20 at 19:55
  • I guess in the end we don't need to establish that the maximum occurs when $a,b,c$ are all non-negative, only that it occurs when all the three factors of $P$ are positive. Then the above proof would be valid. – LHF Jan 24 '20 at 19:58
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Let us prove that the maximum of $P$ is $1600$. Since $P(4,0,1)=1600$, it suffices to prove that $P(a,b,c) \le 1600$.

Since $5c^2 \le 21$, we have $$2c^3 + 5c^2 + 9 > 2c^2\cdot (-3) + 5c^2 + 9 = -c^2 + 9 > 0.$$ Note also that $$(a^3-a^2 + 2)(b^3-b^2+2) = (a+1)(a^2-2a+2)(b+1)(b^2-2b+2).$$ Thus, we only need to prove the case when $(a+1)(b+1) > 0$. We split into two cases:

1) $a+1 > 0$ and $b+1 > 0$: 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:

$$ \begin{aligned}P &\leq \frac{1}{64} \left[(a^2+4)(b^2+4)(c^2+3)\right]^2 \\ &= \frac{1}{64} \left[\frac{1}{25}(a^2+4)\cdot 5(b^2+4)\cdot 5(c^2+3) \right]^2 \\ &\leq \frac{1}{64} \left[\frac{1}{25}\cdot \frac{1}{27} (a^2+5b^2+5c^2+39)^3 \right]^2\\ &= 1600\end{aligned} $$

Note: The conditions $a+1 > 0$ and $b+1 > 0$ are necessary for the first inequality in the chain above when we multiply $a^3-a^2+2\leq \dfrac{1}{8}(a^2+4)^2$ and $b^3-b^2+2\leq \dfrac{1}{8}(b^2+4)^2$.

2) $a+1 < 0$ and $b+1 < 0$: We have \begin{align} &0 < -(a^3-a^2+2) \le \dfrac{121}{544}\left(a^2+\dfrac{37}{11}\right)^2, \\ &0 < -(b^3-b^2+2) \le \dfrac{2}{5}(b^2+1)^2 \end{align} since \begin{align} &\dfrac{121}{544}\left(a^2+\dfrac{37}{11}\right)^2 + (a^3-a^2+2) = \dfrac{1}{544}(121a^2-182a+273)(a+3)^2 \ge 0, \\ &\dfrac{2}{5}(b^2+1)^2 + (b^3-b^2+2) = \dfrac{1}{5}(2b^2-3b+3)(b+2)^2 \ge 0. \end{align} Also, same as in case 1, we have $2c^3 + 5c^2 + 9 \le (c^2+3)^2$. Thus, we have \begin{align} P &\le \dfrac{121}{544}\left(a^2+\dfrac{37}{11}\right)^2 \cdot \dfrac{2}{5}(b^2+1)^2 \cdot (c^2+3)^2\\ &= \dfrac{121}{1360}\left[\left(a^2+\dfrac{37}{11}\right)(b^2+1)(c^2+3)\right]^2\\ &= \dfrac{121}{850000}\left[\left(a^2+\dfrac{37}{11}\right)(5b^2+5)(5c^2+15)\right]^2\\ &\le \dfrac{121}{850000}\Big[\Big(\frac{(a^2+5b^2 + 5c^2) + \tfrac{37}{11} + 5 + 15}{3}\Big)^3\Big]^2\\ &= \dfrac{121}{850000}\Big[\Big(\frac{21 + \tfrac{37}{11} + 5 + 15}{3}\Big)^3\Big]^2\\ &< 1600. \end{align}

We are done.

River Li
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  • Thanks, I need to check your computations on paper, but it looks good at the first read. – LHF Jan 27 '20 at 07:44
  • You may use WA: https://www.wolframalpha.com/ to check them. Moreover, you can simply copy the tex code in my answer (right click on the equation, show math as, tex commands) and paste them on WA. – River Li Jan 27 '20 at 07:52
  • That's what I meant :) I won't do multiplications and divisions literally on paper. But I have to check to understand why you used those bounds for $-(a^3-a^2+2)$ and $-(b^3-b^2+2)$; I suppose it has something to do with the point for which the maximum is reached in case 1. Don't worry though, I'll come back to accept your answer (sometime later this evening). – LHF Jan 27 '20 at 07:56
  • Well, the answer must be reviewed to make sure no errors. – River Li Jan 27 '20 at 08:06
  • I have checked and everything is alright. Thanks! Some observations: For case 2, according to WA, the maximum is around $1260$ for $(a,b,c)=(-3.00141, -1.54865, 0)$. The three separate bounds you used for the three factors are almost equality for $a$ and equality for $c$. The only coarse approximation is for $b$, but not enough to matter. The approximation becomes more significant when you apply AM-GM for those three terms, because they are quite distanced from another. That's why the final value is around $1488$. I'm pretty amazed that such an elementary solution could be found. – LHF Jan 27 '20 at 16:02
  • River Li, can I edit your solution and add the steps you referenced from my answer? I want this thread to have a complete answer for future reference without cross references between our answers. Thanks. – LHF Jan 27 '20 at 16:05
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    @Atticus Just feel free to edit. – River Li Jan 27 '20 at 16:21
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    @Atticus Actually, I first used approximation around $a = -3.00141$ and $b=-1.54865$, but later I found that it is simpler to use $b = -2$ which is enough since there is some space between $1260$ and $1600$. – River Li Jan 27 '20 at 16:25