Find maximum value Given $0 \leq a,b,c \leq \dfrac{3}{2}$ satisfying $a+b+c=3$. Find the maximum value of $$N=a^3+b^3+c^3+4abc.$$
I think the equality does not occur when $a=b=c=1$ as usual. I get stuck in finding the strategy, especially I don't know what to do with $4abc$. Thank you a lot.

For $a=b=\frac{3}{4}$ and $c=\frac{3}{2}$ we get a value $\frac{243}{32}.$
We'll prove that it's a maximal value.
Indeed, $a+b-c=3-2c\geq0,$ which says that there are non-negatives $x$, $y$ and $z$ for which
$a=y+z$, $b=x+z$ and $c=x+y$.
Id est, we need to prove that
$$\sum_{cyc}(x+y)^3+4\prod_{cyc}(x+y)\leq\frac{9(a+b+c)^3}{32}$$ or
$$\sum_{cyc}(x+y)^3+4\prod_{cyc}(x+y)\leq\frac{9(x+y+z)^3}{4}$$ or
$$\sum_{cyc}\left(x^3-x^2y-x^2z+\frac{22}{3}xyz\right)\geq0,$$ which is true by Schur.
