Let's construct a function \begin{align*} f(x) & =1-[x(1-y)+y(1-z)+z(1-x)]\\ & = (y+z-1)x + (yz+1-y-z) \end{align*} The to-be-claimed conclusion is equivalent to showing $f(x) > 0$. Because $f(x)$ is a $1^{st}$ degree polynomial (with respect to $x$) whose graph is a straight line, this above conclusion is equivalent to showing $f(0) > 0$ and $f(1)>0$. This is because if the two ending pointsabove the $x-$axis, the whole segment will be above the $x-$axis. Because $0 < y, z < 1$, we find \begin{align*} f(0)&= yz+1-y-z = (y-1)(z-1) >0\\ f(1)&= (z+y-1)+(yz+1-y-z) = yz > 0 \end{align*} Therefore the conclusion holds.