Since $ x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz))$ we have $ 1-xyz=9-3(xy+yz+zx)$, which is equivalent to $ (x-3)(y-3)(z-3)=-8$, implying that $ x,y,z\in\{-5,-1,1,2,4,5,7,11\}$ for sure. If we have $ -5$, then the others are both $ 2$ or both $ 4$, this gives the solution $ (x,y,z)=(-5,4,4)$ and its permutations. If we don't have $ -5$: we can't have $ -1$, so $ x,y,z\ge 0$ and from the equality case of the power-mean inequality ,we get that $ x=y=z=1$.

W.L.O.G. $ x\geq y\geq z$. If $ x<0$, then $ x^3+y^3+z^3=x+y+z<0$ which contradicts. Since $ x^3\geq 3x-2\Longleftrightarrow (x-1)^2(x+2)\geq 0$, yieilding $ x^3+y^3+z^3\geq 3(x+y+z)-6,$ the eqaulity holds only for $ x=y=z=1.$

I however do think that $ 4^3+4^3+(-5)^3 = 4+4+(-5)=3$... You prove your ineq for $ x$, but you use that $ x\ge-2$. That is true for $ x$ but not for $ y$ and $ z$, so you can't sum them up afterwards.

Just a sidenote: finding all $ x,y,z \in \mathbb Z$ such that $ x^3 + y^3 + z^3 = 3$ is an open problem.

Another Method : from identity $ x^3+y^3+z^3 = (x+y+z)^3-3(x+y)(y+z)(z+x)$ $ 3 = 27-3(x+y)(y+z)(z+x)$ so $ (x+y)(y+z)(z+x) = 8$ thus $ (3-x)(3-y)(3-z) = 8$ which is easy to get $ (x,y,z) = (1,1,1)$

Sorry, It's wrong.

MWIT-19 wrote: $ (3-x)(3-y)(3-z) = 8$ which is easy to get $ (x,y,z) = (1,1,1)$ It's not right because as Peter said, $(x,y,z)=(4,4,-5)$ also works...

Peter wrote: Find all $(x,y,z) \in {\mathbb{Z}}^3$ such that $x^{3}+y^{3}+z^{3}=x+y+z=3$. We observe that $x^3+y^3+z^3=(x+y+z)(x^2+y^2+x^2-xy-yz-zx)+3xyz=x+y+z$ Writing $3$ as $x+y+z$ and factorizing them up we get after a few steps $(x+y+z)^{2}-3(xy+yz+zx) +xyz=1$ Again writing $x+y+z=3$ we get, $3^2 -3(xy +yz+zx)+xyz=1$ =>$8=3(xy+yz+xz)-xyz$ Now factorizing this up we get =>$8=(3-z)xy +3z(x+y)$ now we know $3-z=x+y$ Therefore $8 = (x+y)( xy+3z)$ ,rewriting $(3-z)$ instead of $x+y$ we get =>$8=(3-z)(xy+3z)$ .Now factorize $8$ and find the values of $x$,$y$ and $z$ which we get as $(x,y,z) =(4,4,-5)${and all permutations of it} ;$(1,1,1)$