an equation good to solve

x^4+y^4+z^4\ge x^3+y^3+z^3\\ \ (x^2-y^2)^2+(y^2-z^2)^2+(z^2-x^2)^2<x^4+y^4+z^4
From which we finally have
\boxed{x^4+y^4+z^4<2(x^2y^2+y^2z^2+z^2x^2)}
Suppose none of (x,y,z) is 1.

So
2(\sum {x^2y^2})<\frac{3}{2}(xyz)^2
Thus only value for natural n can be 1...
\boxed{n=1} for x=1, y=2, z=3...

In case any x,y,z equals 1, it is easy to prove that n\le 3
By descent n=2 possibility is ruled out.
\boxed{n=3} for (x,y,z)=(1,1,1)
Thus n=\boxed{1,3}.....