If $P(B\text{ }|\text{ }A)=1-\epsilon$ and $P(C\text{ }|\text{ }B)=1$ then $P(C\text{ }|\text{ }A)\geq 1-\epsilon$

Question

If $$1=P(C\text{ }|\text{ }B)=\frac{P(C\cap B)}{P(B)}$$ then we know that $P(C\cap B)=P(B)$. If $P(B\text{ }|\text{ }A)=1-\epsilon$ for $\epsilon\geq 0$ then $$P(A)=\frac{P(B\cap A)}{1-\epsilon}$$Using this I found that $$P(C\text{ }|\text{ }B)=(1-\epsilon)\frac{P(C\cap A)}{P(B\cap A)}$$ Now I have to show that $$\frac{P(C\cap A)}{P(B\cap A)}\geq 1$$ I don't know how to use the first assumption to show this. I need hints. Thanks.