Just Ineq #1

Define $\displaystyle f(x,y,z) = x^3+y^3+z^3$ with $x+y+z$ fixed . Now it is clear that $f$ is an increasing function and we may observe that,

$\displaystyle f(x+z,y,0)-f(x,y,z)=3xz(z+x)\ge 0$ which in other way round may be written as $f(x,y,z)\le f(x+z,y,0)$ .

It is clear that whenever $z=0$ $f$ attains a maximum which is evident as the last condition forces $x\le x+z\quad\text{\&}\quad z\le 0$ which implies $z=0$ . Thus the problem reduces to maximizing $f(x,y,0)$ with $x+y=3$ .

For $y\le 2$ we have $x\ge 1$ which implies $\dfrac{y}{x}\le 2$ and further $y^3-8x^3\le 0$ with equality iff $y=2x$ .

Since $\displaystyle f(x,y,0)-f(x,2x,0) = y^3-8x^3 \le 0$ which implies $f(x,y,0)\le f(x,2x,0)$ and since we know the maximum is achieved at $y=2x$ we can conclude now that $(x,y,z)=(1,2,0)$ and it's permutations are where maximum is attained.

Therefore $f_{\text{max}}=\boxed{9}$