Does $E\big[\frac{X}{Y}\big]>\beta$ imply $E\big[\frac{X+\beta Z}{Y+Z}\big]>\beta$?

Question

1. Define $X-1\sim Binomial(Y-1,\beta)$. That is, $X$ is the number of successes when the number of trials is $Y$ and the first trial is the success.

2. $Z \sim Binomial\big(\lfloor1+\frac{Y}{X}\rfloor,\beta\big) $

Under this setting, does $E\big[\frac{X}{Y}\big]>\beta$ imply that $E\big[\frac{X+\beta Z}{Y+Z}\big]>\beta$?

Intuitively, I think that $\frac{X+\beta Z}{Y+Z}$ is the weighted average between $\frac{X}{Y}$ and $\beta$. Therefore, my prediction is that $E\big[\frac{X}{Y}\big]>E\big[\frac{X+\beta Z}{Y+Z}\big]>\beta$ holds.

Any comments or inputs would be appreciated.