<s>[INST] 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 
 [/INST] Here is how I would solve it, First, I would eliminate x.
\begin{eqnarray*}
x &=& 3 - y - z \\
f(x,y,z) &=& (3 - y - z)^3 + y^3 + z^3
\end{eqnarray*}
then I would apply the second derivative test. More information about the second derivative test can be found at the following URL:
http://faculty.csuci.edu/brian.sittinger/2nd_DerivTest.pdf
First, we find the partial derivatives:
\begin{eqnarray*}
f_y &=& -3(3 - y - z)^2 + 3y^2
f_z &=& -3(3 - y - z)^2 + 3z^2
f_yz &=& 6(3 - y - z) + 6y \\
f_yy &=& -6(3 - y - z) + 6y \\
f_zz &=& -6(3 - y - z) + 6z
\end{eqnarray*}
Now, we find the critical points:
\begin{eqnarray*}
-3(3 - y - z)^2 + 3y^2 &=& 0 \\
-3(3 - y - z)^2 + 3z^2 &=& 0 \\
3y^2 - 3z^2 &=& 0 \\
y^2 &=& z^2 \\
\end{eqnarray*}
Therefore, $y = z = 0$ is a critical point. However, it is a minimum not a maximum. We also have an infinity number of critical points, so I am not
sure how to proceed.
Bob
</s>