4. Proof: Let
$$\begin{array}{r} A=x^{k+1}+y^{k}+z^{k}=x^{k+1}+\left(y^{k}+z^{k}\right)(x+y \\ +z)=x^{k+1}+y^{k+1}+z^{k+1}+x\left(y^{k}+z^{k}\right)+y^{k} z+ \end{array}$$
$y z^{k}$,
$$\begin{array}{r} B=y^{k+1}+x^{k}+z^{k}=y^{k+1}+\left(x^{k}+z^{k}\right)(x+y \\ +z)=x^{k+1}+y^{k+1}+z^{k+1}+y\left(x^{k}+z^{k}\right)+x^{k} z+ \end{array}$$
$x z^{k}$,
$$\begin{array}{r} C=z^{k+1}+y^{k}+x^{k}=z^{k+1}+\left(y^{k}+x^{k}\right)(x+y \\ +z)=x^{k+1}+y^{k+1}+z^{k+1}+z\left(y^{k}+x^{k}\right)+x^{k} y+ \end{array}$$
$x y^{k}$,

By symmetry, assume $x \geqslant y \geqslant z$, then $A \leqslant B \leqslant$ $C, \frac{x^{k+2}}{A} \geqslant \frac{y^{k+2}}{B} \geqslant \frac{z^{k+2}}{C}$,
By Chebyshev's inequality,
$$\begin{array}{l} \left(\frac{x^{k+2}}{A}+\frac{y^{k+2}}{B}+\frac{z^{k+2}}{C}\right)(A+B+C) \geq \\ 3\left(x^{k+2}+y^{k+2}+z^{k+2}\right), \end{array}$$

So, the left side $\geqslant \frac{3\left(x^{k+2}+y^{k+2}+z^{k+2}\right)}{A+B+C}$,
Also, $A+B+C=3\left(x^{k+1}+y^{k+1}+z^{k+1}\right)+$ $2\left(x^{k} y+y^{k} z+z^{k} x\right)+2\left(x y^{k}+y z^{k}+z x^{k}\right) \leqslant$ $7\left(x^{k+1}+y^{k+1}+z^{k+1}\right)$,

So, the left side $\geqslant \frac{3\left(x^{k+2}+y^{k+2}+z^{k+2}\right)}{7\left(x^{k+1}+y^{k+1}+z^{k+1}\right)} \geqslant$ $\frac{\left(x^{k+1}+y^{k+1}+z^{k+1}\right)(x+y+z)}{7\left(x^{k+1}+y^{k+1}+z^{k+1}\right)}=\frac{1}{7}$.