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.

  • $\begingroup$ I immediately thought multi-variable calculus maximization problem, but I really like this solution (even though I don't exactly understand the intuition or symbols) because it seems much simpler. $\endgroup$ – Christopher Marley Oct 18 '18 at 5:20

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 $c=3/2, a=b$.


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