Inequality on distances of probabilities

Question

$0 < x, y, z < 1$

Assume

$\frac{y}{2} < z$

Can we conclude that:

$\vert\frac{x}{2} -\frac{y}{2}\vert < \vert x-z\vert$

Answer 1

No.

Let $X=Z=\frac{9}{10},Y=\frac{1}{10}$.

RHS of the inequality is zero, LHS is not, therefore this is wrong.