Suppose $X>0$ and $Y>0$ are not necessarily independent random variables
Another random variable is $Z>0$
Then, does $E\big[\frac{1+Y}{1+X}\big]>\beta$ imply $E\big[\frac{1+Y+\beta Z}{1+X+Z}\big]>\beta$?
As an additional question, suppose there is a constraint: $X>Y>0$
Then, does $1>E\big[\frac{1+Y}{1+X}\big]>\beta$ imply $E\big[\frac{1+Y+\beta Z}{1+X+Z}\big]>\beta$?
No. Suppose that $(X,Y)$ takes on the values $(1,2)$ and $(6,3)$ with equal probability, $Z=1$ always, and $\beta=1$. Then
$$
\mathbb E\left[\frac{1+Y}{1+X}\right] = \frac{\frac{1+2}{1+1} + \frac{1+3}{1+6}}{2} = \frac{29}{28} > 1
$$
but
$$
\mathbb E\left[\frac{1+Y + \beta Z}{1+X + Z}\right] = \frac{\frac{1+2+1}{1+1+1} + \frac{1+3+1}{1+6+1}}{2} = \frac{47}{48} < 1.
$$
(Method: I chose $Z\equiv 1$, $\beta=1$, and $(X,Y)$ to be uniform on $\{(x_1,y_1), (x_2, y_2)\}$ to simplify things, then used Mathematica's FindInstance command to find values of $x_1, x_2, y_1, y_2$ that satisfy the inequalities above.)
We can answer the extra question with the same method, but taking $\beta=\frac12$ and $(X,Y)$ to be $(14,3)$ or $(3,2)$ with equal probability.
