How do I find the maximum value of $a^2(b-c)+b^2(c-b)+c^2(1-c)$, if $0\le{a}\le{b}\le{c}\le{1}$?
I have tried plugging in values, tried manipulating and inequality by means but that didn't worked. How do I approach this question?
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2$\begingroup$ What else have you tried? EG Can you show that $ a = 0 $ at the maximum? That would reduce to a 2 variable inequality, which might be easier to manage. $\endgroup$– Calvin LinFeb 18 at 13:09
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$\begingroup$ I would start by observing that the expression is a sum of two positive terms: $(a^2-b^2)(b-c)+c^2(1-c)$. The last term reaches a maximum of $\frac{4}{27}$ when $c=\frac 23$. Then focus to determine the maximum of $(a^2-b^2)(b-\frac 23), a,b\lt \frac 23$. $\endgroup$– WindSoulFeb 18 at 13:41
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$\begingroup$ @WindSoul I don't think that's a good suggestion. $\max [f(x) + g(x] \leq \max f(x) + \max g(x)$, and these do not need to occur at the same point. $\endgroup$– Calvin LinFeb 19 at 13:37
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$\begingroup$ I think that was an excellent suggestion, not because is right but because is accompanied by another suggestion which proved at the time to be the only valid start point in the answer: the fact that the expression is the sum of two positive terms. While developing the answer I realized my assumption was wrong but the perspective on answer was in that comment which at the time was the only response to the OP. I may have lost the battle with time at being the first to provide the answer but I was sufficiently right to be able. At least one other answer starts exactly from the same point. $\endgroup$– WindSoulFeb 19 at 15:13
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3 Answers
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$$\underbrace{(c-b)}_{\ge0}\underbrace{(b^2-a^2)}_{\ge0}+c^2\underbrace{(1-c)}_{\ge0}\\ $$ $b^2-a^2$ is the only term dependent on $a$ and thus it attains a maximum at $a=0.$ The problem then reduces to $$b^2(c-b)+c^2(1-c)$$ for $0\le b\le c\le1.$ Sub $b=rc.$ $$r^2c^3(1-r)+c^2(1-c)\\ c^2(r^2(1-r)c+1-c)$$ $r^2(1-r)\overset{\partial_r}{\longrightarrow}r(2-3r)$ has a maximum at $r=\frac23.$ We then maximize $$c^2\left(1-\frac{23}{27}c\right)\overset{\partial_c}{\longrightarrow}c\left(2-\frac{23}9c\right)\\ c=\frac{18}{23}\\ b=\frac23\cdot\frac{18}{23}=\frac{12}{23}$$ The maximum is thus $$\frac{108}{529}.$$
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Capitalizing on writing the expression as a sum of two positive numbers, indicates the value of a:
$$\begin{align}E(a,b,c)&=\underbrace{(b^2-a^2)}_{\ge 0}\underbrace{(c-b)}_{\ge 0}+\underbrace{c^2(1-c)}_{\ge 0}, a,b,c \lt 1\\ &\le (b^2-0)(c-b)+c^2(1-c), \boxed{a= 0}\end{align}$$
The first term $b^2(c-b)$ could be analyzed using $f(x)=x^2(c-x), x\lt c$. This cubic has a maximum at $x=\frac 23 c$, therefore $\boxed{b=\frac 23 c}$
Replacing $b=\frac 23 c$, the original expression after considering $a=0$ becomes:
$$E(a,b,c)\lt \frac {4}{27}c^3+c^2(1-c), E(a,b,c)\lt -\frac{23}{27}c^3+c^2$$
The function $g(x)=-\frac{23}{27}x^3+x^2$ has a maximum at $x=\frac{18}{23}$. Therefore $\boxed{c=\frac{18}{23}}$.
Replacing $a,b,c$ finds the maximum of the original expression.
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If you're interested in a quick answer, you can use quantifier elimination, for example in Mathematica:
$$\forall_{a,b,c,0\leq a\leq b\leq c\leq 1}\qquad a^2(b-c)+b^2(c-b)+c^2(1-c)<m$$
Then use Resolve[%, Reals] to obtain:
$$m>\frac{108}{529}.$$
Similarly, using Maximize, you can also obtain the values $a=0$, $b=\frac{12}{23}$, and $c=\frac{18}{23}$.
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1$\begingroup$ Thank you for your answer. It exemplifies how important is to being able to use tools to reach numbers that are beyond mental number crunching. Would I have had the same skill, maybe I would not have been wasting time in my answer with the correct determination of c. $\endgroup$– WindSoulFeb 19 at 15:25