<s>[INST] $\max a^3+b^3+c^3+4abc$ sub $0\leq a,b,c \le 3/2$ and $a+b+c=3$ 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 
$$
\max_{(a,b,c) \in S} a^3+b^3+c^3+4abc.
$$
 [/INST] 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:


*

*If $a > \frac97$, $(a', b', c')$ is better than $(a, b, c)$.

*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:


*

*We know that we can increase $a^3 + b^3 + c^3 +4abc$ by replacing $(a, b, c)$ by $(a, \frac32, \frac32-a)$.

*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...
</s>