A gambler went to a casino where one can stake any amount against a chance of probability $p$ of winning a prize equal to his stake, where $0 < p < \frac12$. Also, the gambler has to pay an income tax at a rate of $t \in (0,1)$ on the positive income, that is, if the gambler stake an amount of $10$ units, then the probability of winning a prize of $10$ units is $p$ and the tax he has to pay is $10t$ units.
Suppose that the gambler has $x$ units and wants to win $y$ additional units. Now assume that, the probability for the gambler to win $y$ units with an initial fortune of $x$ units is $q$. Then I want to prove that,
$$ q \le \frac{(1-t)x}{(1-t)x+y} $$
My tried solution
Basically there is two cases.
Let $y \le x(1-t)$. Then, $\frac{y}{1-t}\le x$. So, in this case, to win additional $y$ units, the gambler can stake off amount $\frac{y}{1-t}$ units. Then with probability $p$ he will receive a prize of $\frac{y}{1-t}$ units minus the tax deduction of $\frac{yt}{1-t}$ units, so total $y$ units. Now notice that, $$y \le x(1-t) \implies \frac{(1-t)x}{(1-t)x+y} \ge \frac{1}2>p.$$ Since in this case $p=q$, so we are done. Now move to case-$2$.
Let $y \ge x(1-t)$. Now I stuck here. I am unable to show the required inequality in this case. Please help me to solve this.