# Inequality with probability

I am stuck with some inequality.Let $A,B$ and $C$ be events in a probability space $(\Omega,\Sigma, \mu)$ where $B\subset C$. Define $r=\mu(A\mid C)$ and $x=\mu(A\mid B)$. Assume $\mu(B\mid C)\geq p$ for some $p\in[0,1]$. I want to show the following inequality $xp\leq r\leq 1+x-p$.

What I did so far is the following:

$$x=\frac{\mu(A\cap B)}{\mu(B)}=\frac{\mu(A\cap C)}{\mu(B)}-\frac{\mu(A\cap [C\setminus B])}{\mu(B)}\\ =r\frac{\mu(C)}{\mu(B)}-\frac{\mu(A\cap [C\setminus B])}{\mu(B)}.$$

Now from here we have $x\leq r\frac{\mu(C)}{\mu(B)}\implies px\leq r$. I could not prove the second part of the inequality. Any help is greatly appreciated.

For the second part of the inequality, write $$\mu(B\cap C)-\mu(A'\cap C)\stackrel{(1)}\le\mu(B)-\mu(A'\cap B)=\mu(A\cap B)$$ where $A'$ denotes the complement of $A$. In (1) we use the fact that $B\subset C$. Now divide through by $\mu(C)$ to obtain $$\mu(B\mid C)-\mu(A'\mid C)\le {\mu(A\cap B)\over \mu(C)}\stackrel{(2)}\le{\mu(A\cap B)\over \mu(B)}=\mu(A\mid B)$$ where (2) again uses $B\subset C$. Since $\mu(B\mid C)\ge p$, this gives $$p-(1-r)\le x\;.$$