Inequality for conditional probabilities

Question

does somebody have an idea how to prove this proposition:

$A$ arbitrary, $B\subset C$ and $P(B)>0$ => $P(A|C)\ge P(A|B)$. It should be simple but somehow I cannot get it.

Thanks in advance.

Answer 1

This is not true. Let $B = A, C = \Omega$, then $$ \mathbb{P}(A|C) = \mathbb{P}(A|\Omega) = \mathbb{P}(A) \leq 1 = \mathbb{P}(A|A) = \mathbb{P}(A|B). $$

Answer 2

I don't think it's true. Let $A=B.$