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 $\alpha = \infty$.

Recall that the support $S(T)$ of a real valued random variable $T$ is the set of real numbers $t$ such that the events $[|T-t|\le\varepsilon]$ have positive probability for every positive real number $\varepsilon$.
If $Y\le X$ almost surely, that is, if $\sup S(Y)\le\inf S(X)$, then every nonnegative real number $u$ works since $Y-X(1+Z)\le Y-X\le0$ and $Z\ge0$ almost surely. 
Otherwise, that is, if $\sup S(Y)>\inf S(X)$, the real number $u$ works if and only if $u\ge u_*$ with
$$
u_*=\frac{\sup S(Y)-\inf S(X)}{\inf S(Z)}-\inf S(X).
$$
To prove this, one optimizes first in $Y$, then in $X$ and finally in $Z$. These three operations can be performed separately because $X$, $Y$ and $Z$ are independent.
In particular, $u_*$ is finite if and only if the ratio defining $u_*$ is finite if and only if the numerator of this ratio is finite and its denominator is positive, that is, if and only if $S(Y)$ is bounded above by a finite $y$ and $S(Z)$ is bounded below by a positive $z$, that is, $[Y\le y]$ and $[Z\ge z]$ both have full probability. Otherwise, $u_*=+\infty$, which means that no value of $u$ is suitable.
