Probability inequality Let $X,Y,Z$ be non-negative independent r.v. I should find for which $u\geq 0$ holds
$$
Zu\geq Y - X(1+Z)
$$
with probability one. Clearly, if $P(Z = 0)>0$ then  the value of $u$ doesn't matter and we should have $X\geq Y$ with probability one.
The question is the following: what if $P(Z = 0) = 0$? Could we divide an inequality by $Z$ and continue working?
If yes I expect an answer like
$$
u\geq \alpha
$$
for 
$$
\alpha = \inf\left(a:P\left[\frac{Y-X}{Z}-X\leq a \right]= 1\right)
$$
with possibly $\alpha = \infty$.

Since
$$
P[Zu \ge Y - X(1 + Z)] = P[Zu \ge Y - X(1 + Z),Z > 0] + P[Zu \ge Y - X(1 + Z),Z = 0],
$$
$P(Z=0)=0$ yields
$$
P[Zu \ge Y - X(1 + Z)] = P[Zu \ge Y - X(1 + Z),Z > 0] = P\bigg[u \ge \frac{{Y - X}}{Z} - X,Z > 0\bigg].
$$
