$\frac{1}{x}+\frac{1}{y} + \frac{1}{z} \geq x+y+z\Leftrightarrow xy+xz+yz-x-y-z\geq0\Leftrightarrow(1-x)(1-y)(1-z)\geq0.$ If $1-x\geq0,$ $1-y\geq0,$ $1-z\geq0$ then $\{x,y,z\}\subset(0,1].$ Hence, $\{x^k,y^k,z^k\}\subset(0,1].$ Hence, $(1-x^k)(1-y^k)(1-z^k)\geq0.$ Id est, $\frac{1}{x^{k}}+\frac{1}{y^{k}} + \frac{1}{z^{k}} \geq x^{k}+y^{k}+z^{k}.$ If $1-x\leq0,$ $1-y\leq0$ and $1-z\geq0$ then $1-x^k\leq0,$ $1-y^k\leq0,$ $1-z^k\geq0.$ Id est, $(1-x^k)(1-y^k)(1-z^k)\geq0$ and $\frac{1}{x^{k}}+\frac{1}{y^{k}} + \frac{1}{z^{k}} \geq x^{k}+y^{k}+z^{k}.$

Nice solution One can also consider the function : $f(t)=(t-x)(t-y)(t-z)$. Then $f(1)=(1-x)(1-y)(1-z) \geq 0$. Since $x,y,z$ are positive, we have $s(1-x^{2k}) = s(1-x^{k})= s(1-x)$ (where $s(n)$ denotes the sign of n), so $(1-x^{2k})(1-y^{2k})(1-z^{2k}) \geq 0$. Dividing both sides by $(xyz)^{k}$ we'll get the desired result from AM-GM inequality.