<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] You have correctly established that $z=0$. From there you have $y=3-x$ so substitute that into $f$. As $y\le2$ then $1\le x\le2$.
$$f(x,3-x,0)=x^3+(3-x)^3=9x^2+27x+27=9(x^2+3x+3)$$
$$=9\left(x-\frac{3}{2}\right)^2+\frac{27}{4}$$
This quadratic has minimum at $x=\frac{3}{2}$ and the maximum is only limited by the domain of $x$ which leads to the answer of $x=1$ or $x=2$ so the three numbers are $0,1,2$ and the maximum is $9$.
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