<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] Hint : Use : 
$$(x+y+z)^3=x^3+y^3+z^3+3(x+y)(y+z)(z+x)$$
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