Just "simple" Ineq #2

Let $f(x,y,z)=x^3+y^3+z^3+4xyz$ with the constraints $0\le x,y,z \le \dfrac{3}{2}$ and $x+y+z=3$ . It is needless saying that the function is increasing although we may observe that , $f(x,y,z)-f(x,y,x+y) = z^3 -(x+y)^3 + 4xy(z-x-y)$ . Now if we assume $z\le x+y$ then it follows that $f(x,y,z)\le f(x,y,x+y)$ which again establishes that $z\le x+y$ must hold. So we get $z=\dfrac{3}{2}$ and left with the constraint $x+y=\dfrac{3}{2}$ to maximize $f(x,y,3/2)$ .

Since $f(x,y,3/2)=\dfrac{27}{8}+x^3+y^3+6xy$ with $2(x+y)=3$ we would show that the maximum is achieved at $x=y$ . It is easy to see that $f(x,y,3/2)=\dfrac{27}{8}+x^3+y^3+6xy=\dfrac{27}{4}+\dfrac{3xy}{2} \le \dfrac{27}{4}+\dfrac{3}{8}(x+y)^2 =\dfrac{243}{32}$ where equality holds iff $x=y=\dfrac{3}{4}$ . Thus we conclude that $f_{\text{max}} = \dfrac{243}{32}\approx \boxed{7.59375}$ at $(x,y,z)=\left(\dfrac{3}{4},\dfrac{3}{4},\dfrac{3}{2}\right)$ .