$\max a^3+b^3+c^3+4abc$ sub $0\leq a,b,c \le 3/2$ and $a+b+c=3$

Question

Let $S$ be the set of $(a,b,c) \in \mathbb{R}^3$ such that $0\leq a,b,c \leq \frac{3}{2}$ and $a+b+c=3$. Find $$ \max_{(a,b,c) \in S} a^3+b^3+c^3+4abc. $$

Answer 1

Let us start with a triple $(a, b, c)$. Let us try to "tune" the triple and see if we can come out with a candidate $(a', b', c')$ with a larger $a'^3 + b'^3 + c'^3 + 4a'b'c'$. Heuristic tells us that $(a', b', c') = (a, \frac{b+c}{2}, \frac{b+c}{2})$ worth some consideration.

Let $A = a^3 + b^3 + c^3 + 4abc$ and $B = a'^3 + b'^3 + c'^3 + 4a'b'c'$. By some calculation

$$ \begin{aligned} A - B &= \left(b^3 + c^3 - \frac{(b+c)^3}{4}\right) + 4a\left(bc - \frac{(b+c)^2}{4}\right)\\ &= (b+c)\left(\frac{3(b-c)^2}{4}\right) - a(b-c)^2 = (b-c)^2\left(\frac{3(b+c)}{4}-a\right)\\ &= (b-c)^2 \left(\frac{3(a+b+c)}{4}- \frac74 a\right) = (b-c)^2\left(\frac{9-7a}{4}\right). \end{aligned} $$

This has two implications:

1. If $a > \frac97$, $(a', b', c')$ is better than $(a, b, c)$.
2. If $a < \frac97$, $(a, b, c)$ is better than $(a', b', c')$. Moreover, by looking at the difference, we see that the best we can do to maximize $a^3 + b^3 + c^3+ 4abc$ fixing $a$ is to maximize the difference of $b$ and $c$.

Now, from a set $(a, b, c)$, what we can do is as the follows.

By swapping the variables if necessary, we can assume $a \leq 1$. Let us do the follows:

1. We know that we can increase $a^3 + b^3 + c^3 +4abc$ by replacing $(a, b, c)$ by $(a, \frac32, \frac32-a)$.
2. We also know that we can increase $a^3 + b^3 + c^3 +4abc$ by replacing $(a, \frac32, \frac32-a)$ by $(\frac 32, \frac34, \frac 34)$. From this point we can do no more, hence the maximal value is

$$ \left(\frac32\right)^3 + \left(\frac34\right)^3 + \left(\frac34\right)^3 + 4\times \frac32 \times \frac 34 \times \frac 34 = \frac{243}{32} \approx 7.59. $$

This is indeed suprising...

Answer 2

I don't know a nice solution. However, the gradient of $f(a,b,c)=a^3+b^3+c^3+4abc-\lambda(a+b+c-3)$ is $$ \nabla f(a,b,c,\lambda)=\left(3a^2+4bc-\lambda , 3b^2+4ac-\lambda , 3c^2+4ab-\lambda , 3-a-b-c\right). $$ By Weierstrass' theorem, the maximum exists on, let's say, $(x,y,z,w)$. Hence, a necessary condition is $\nabla f(x,y,z,w)=(0,0,0,0)$. From the first two conditions here we get $$ 3x^2+4yz-w =3y^2+4 xz-w \implies (x-y)(3x+3y-4z)=0\,\,\,\,(\star) $$ and the other two cyclic conditions.

We claim that at least two variables need to be equal. This is clear in the case that $x-y=0$ (resp. $y=z$ ro $z=x$) in $(\star)$. Otherwise we have wlog $3x+3y-4z=3y+3z-4x$, which implies $x=z$ (and similarly the others). Hence, we reduce to the two variable problem: maximize $$ g(x,y)=2x^3+y^3+4x^2y\,\,\, \text{ sub } 2x+y=3 \text{ and }0\le x,y \le \frac{3}{2}. $$ The condition $0\le y=3-2x \le \frac{3}{2}$ implies that, substituing $y=3-2x$, we rewrite the problem as to maximize, under the condition $3/4 \le x \le 3/2$, $$ h(x):=g(x,3-2x)=2x^3+(3-2x)^3+4x^2(3-2x) \implies h^\prime(x)=6(-7x^2+16x-9). $$ Considering that $-7x^2+16x-9$ has the maximum in $8/7$ and has a $0$ in $1$, the other zero is in $9/7$, which is smaller than $3/2$. On the other hand, $h(3/4)$ is greater than $h(9/7)$, hence the maximum of the function is at the point $(3/4, 3/4, 3/2)$, that is, $$ 2\left(\frac{3}{4}\right)^3+\left(\frac{3}{2}\right)^3+4\cdot \frac{3^3}{32}=\frac{243}{32}. $$