$\max a^3+b^3+c^3+4abc$ 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 
$$
\max_{(a,b,c) \in S} a^3+b^3+c^3+4abc.
$$

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,
$$
2\left(\frac{3}{4}\right)^3+\left(\frac{3}{2}\right)^3+4\cdot \frac{3^3}{32}=\frac{243}{32}.
$$
