Prove the inequality with $x, y,z$ is the sides of a triangle and $n\in \mathbb Z \land n\ge2$ $${x^n}y(x - y) + {y^n}z(y - z) + {z^n}x(z - x) \ge 0 \tag 1$$
I can prove the inequality with $n=2$: $$(1)\iff x(y-z)^2(x+z-x) + y(x-y)(x-z)(x+y-z) \ge 0$$ or apply Ravi Substitution: $$(1)\iff \frac {a^2}b+\frac {b^2}c + \frac {c^2}a \ge a+b+c\ $$ (alway true, by AM-GM)
But I can't prove the general problem above
Please forgive my grammar
Notice that $$x^ny(x - y) + y^nz(y - z) + z^nx(z - x)=\left|\begin{matrix}x^ny & y^nz & z^nx\\1 & 1 & 1\\z & y & x\end{matrix}\right|=(x^ny,y^nz,z^nx)\cdot\left((1,1,1)\times(z,y,x)\right)$$ Without loss of generality, let $x\geqslant y\geqslant z>1$, for otherwise we can put a scalar on $x,y,z$. Then by drawing the vectors $(x^ny,y^nz,z^nx),(1,1,1),(z,y,x)$ you can see they are in right-hand order and hence the determinant is nonnegative.
you can also induct further: if $n=k$ is OK,then:$n=k+1$, WLOG,there is two cases:
case I: $x\ge z \ge y$,
${x^{k+1}}y(x - y) + {y^{k+1}}z(y - z) + {z^{k+1}}x(z - x) \ge z{x^{k}}y(x - y)+z{y^{k}}z(y - z)+ z{z^{k}}x(z - x)=z\left({x^{k}}y(x - y) + {y^{k}}z(y - z) + {z^{k}}x(z - x)\right) \ge 0$
case II: $x\ge y \ge z$
${x^{k+1}}y(x - y) + {y^{k+1}}z(y - z) + {z^{k+1}}x(z - x) \ge z{x^{k}}y(x - y)+z{y^{k}}z(y - z)+ z{z^{k}}x(z - x)=z\left({x^{k}}y(x - y) + {y^{k}}z(y - z) + {z^{k}}x(z - x)\right) \ge 0$
