Maximum of $x^3+y^3+z^3$ with $x+y+z=3$ It is given that, $x+y+z=3\quad 0\le x, y, z \le 2$ and we are to maximise $x^3+y^3+z^3$. 
My attempt : if we define $f(x, y, z) =x^3+y^3 +z^3$ with $x+y+z=3$ it can be shown that, 
$f(x+z, y, 0)-f(x,y,z)=3xz(x+z)\ge 0$ and thus $f(x, y, z) \le f(x+z, y, 0)$. This implies that $f$ attains it's maximum whenever $z=0$. (Is this conclusion correct? I have doubt here). 
So the problem reduces to maximise $f(x, y, 0)$ which again can be shown that $f(x, y, 0)\le f(x, 2x,0)$ and this completes the proof with maximum of $9$ and equality at $(1,2,0)$ and it's permutations. 
Is it correct? I strongly believe even it might have faults there must be a similar way and I might have made mistakes. Every help is appreciated 

$$
\begin{eqnarray}
&(x+y+z)^3  &=&  x^3 + y^3 + z^3 + 3(x+y)(y+z)(z+x)\\
\implies & x^3 + y^3 + z^3  &=&  (x+y+z)^3 - 3(x+y)(y+z)(z+x)\\
&& = & 27 - 3(x+y)(y+z)(z+x)
\end{eqnarray}
$$
Now, $x^3+y^3+z^3$ is maximum when $t = (x+y)(y+z)(z+x)$ is minimum. Now since $x$, $y$ and $z$ are each non-negative, therefore $t$ is non-negative. Also, $x,\,y,\,z \in [0,\,2]$. So, $t$ takes minimum value when the variables take values $0,\,1,\,2$. So, $t_\text{min} = (0+1)(1+2)(2+0) = 6$.
So $\max (x^3+y^3+z^3)=27-3\times6=9$.
