I failed miserably to prove this by induction. But then I realised that the condition is equivalent to $(1-x)(1-y)(1-z) \ge 0$ So either all of $x,y,z$ are between $0$ and $1$, or one of them is and the other two are $> 1$ We need to prove that $(1-x^k)(1-y^k)(1-z^k) \ge 0$ But if $0<x<1$ then $0<x^k <1$. And if $x>1$ then $x^k>1$, and similarly for $y$ and $z$ So now it's obvious