If $P(A) \neq 0$ and $P(B) \neq 0$, then $P(B|A) \geq P(B)$ is equivalent to $P(A|B) \geq P(A)$ I am puzzled by the intuition behind the following fact:
If $P(A) \neq 0$ and $P(B) \neq 0$, then $P(B|A) \geq P(B)$ is equivalent to $P(A|B) \geq P(A)$. 
This is easy enough to show by definition of conditional probability, but I would like to have some sort of geometric intuition behind this. 
I can create positive example pictures, but not one that shows me why this statement must hold.
I would greatly appreciate some help. Thanks

Both statements are saying that $P(A \cap B) \ge P(A)\cdot P(B)$. Note that $P(A)\cdot P(B)$ corresponds to $P(A \cap B)$ if $A$ and $B$ were independent. Thus $P(A \cap B) \ge P(A)\cdot P(B)$ means that there is some positive correlation (in a figurative sense) between these two events.
On the other hand, the phrase that always pops in my mind when I am talking about conditional probability is you are changing your universe. $P(X|Y)$ means that you are looking at $P(X)$ in a different universe, i.e. $Y$. Since there is a positive correlation between these two, it makes sense that $P(A|B) \ge P(A)$. This is because we are changing our universe to $B$, and since we know that $B$ has a positive correlation with $A$, it means that in this new universe, we are more likely to see $A$. This is symmetric in $A$ and $B$ of course.
I don't know if this is intuition enough.
