I am trying to prove a strange inequality regarding prior probabilities.
A large number of patients claim to have contracted a disease. Let $D$ be the event that this story is true. Now let $R$ be the event that a large number of patients report it to be true. Suppose that $P(R|D)=1$ (this is obvious since if you have the disease it makes sense to always report it) and $P(R|D^c)=\beta$ where $\beta\in[0,1)$. I am trying to prove that $P(D)>P(D^c)$ whenever $P(D)>\frac{\beta}{1+\beta}$.
Attempt: $1=P(R|D)=\frac{P(R,D)}{P(D)}\le \frac{P(R)}{P(D)}$. Hence $P(R)=P(D)$, and so $\beta=0$, while $P(D)=1>P(D^c)$. Obviously I didnt use the main inequality given to me, so this is likely incorrect. But what part of the argument is not right?
You wrote $$1 = \Pr[R \mid D] = \frac{\Pr[R \cap D]}{\Pr[D]} \color{red}{\le} \frac{\Pr[R]}{\Pr[D]}$$ and then incorrectly inferred $\Pr[R] = \Pr[D]$. Because you used an inequality, the correct conclusion is $$\Pr[R] \ge \Pr[D].$$