Solution:

The given expression is symmetric, so without loss of generality we may assume that $x \geqslant y \geqslant z$. Then
$$x^{k+1}+y^{k}+z^{k} \leqslant y^{k+1}+z^{k}+x^{k} \leqslant z^{k+1}+x^{k}+y^{k}$$
Indeed, it suffices to prove the first inequality, i.e. that $x^{k+1}+y^{k} \leqslant y^{k+1}+x^{k}$. This inequality is equivalent to $\left(\frac{y}{x}\right)^{k} \leqslant \frac{1-x}{1-y}$. Since $y \leqslant x$, it suffices to prove that $\frac{y}{x} \leqslant \frac{1-x}{1-y}$, which is equivalent to the true inequality $0 \leqslant x-x^{2}-y+y^{2}=(x-y)(1-x-y)=(x-y) z$.

By applying Chebyshev's inequality to the triples $\left(x^{k+2}, y^{k+2}, z^{k+2}\right)$ and $\left(\frac{1}{x^{k+1}+y^{k}+z^{k}}, \frac{1}{y^{k+1}+z^{k}+x^{k}}, \frac{1}{z^{k+1}+x^{k}+y^{k}}\right)$ we obtain
$$\sum_{\text{cyc}} \frac{x^{k+2}}{x^{k+1}+y^{k}+z^{k}} \geqslant \frac{1}{3} \sum_{\text{cyc}} x^{k+2} \sum_{\text{cyc}} \frac{1}{x^{k+1}+y^{k}+z^{k}} = L$$
If in $L$ we apply Chebyshev's inequality once more to the triples $(x, y, z)$ and $\left(x^{k+1}, y^{k+1}, z^{k+1}\right)$ we obtain
$$L \geqslant \frac{1}{3} \cdot \frac{1}{3} \sum_{\text{cyc}} x \sum_{\text{cyc}} x^{k+1} \sum_{\text{cyc}} \frac{1}{x^{k+1}+y^{k}+z^{k}} = L'$$
From the Cauchy-Schwarz-Bunyakovsky inequality it follows that
$$\sum_{\text{cyc}} \frac{1}{x^{k+1}+y^{k}+z^{k}} \sum_{\text{cyc}}\left(x^{k+1}+y^{k}+z^{k}\right) \geqslant 9$$
so
$$L' \geqslant \frac{x^{k+1}+y^{k+1}+z^{k+1}}{x^{k+1}+y^{k+1}+z^{k+1}+2\left(x^{k}+y^{k}+z^{k}\right)}$$
and hence it suffices to prove that
$$3\left(x^{k+1}+y^{k+1}+z^{k+1}\right) \geqslant x^{k}+y^{k}+z^{k}$$
The last inequality is obtained by applying Chebyshev's inequality once more to the triples $(x, y, z)$ and $\left(x^{k}, y^{k}, z^{k}\right)$.

Equality in all the inequalities applied holds if and only if $x=y=z$, i.e. if and only if $x=y=z=\frac{1}{3}$.