# 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 $$\frac{243}{32}.$$
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.
This seems easier using multiple variable calculus. Substitute $$c=3-a-b$$ to get $$N=27-27a-27b+9a^2+30ab+9b^2-7a^2b-7ab^2$$. There are 4 critical points, the maximum value at the critical points is 7.16.. So now look on the boundary where $$a=3/2$$, then $$N=27/4+(9/4)b-(3/2)b^2$$ has a critical point when $$b=3/4$$ giving the value $$N=243/32=7.59..$$. The maximum also occurs $$b=3/2, a=c$$ or $$c=3/2, a=b$$.