True or False : $P( B \mid A \cup B) \geq P(B \mid A)$ Trying to prove whether this is T or F and I get stuck..
This is my process so far:
$P( B \mid A \cup B) \geq P(B \mid A)$ 
$=\frac{P(B\cap(A\cup B))}{P(A \cup B)} \geq \frac{P(B \cap A)}{P(A)}$
$=\frac{P(B)}{P(A \cup B)}\geq \frac{P(B \cap A)}{P(A)}$
 A: More intuitive way per OPs request:
Take the reciprocal of your last inequality:
$$\frac{P(A \cup B)}{P(B)}\leq \frac{P(A)}{P(B \cap A)}$$
Then subtract $1$ both sides"
$$\frac{P(A \cup B)-P(B)}{P(B)}\leq \frac{P(A)-P(B \cap A)}{P(B \cap A)}$$
That can be rewritten as:
$$\frac{P(A \setminus B)}{P(B)}\leq \frac{P(A \setminus B)}{P(B \cap A)}$$
Which is obvious
A: Note that
$$
\frac{P(B)}{P(A\cup B)}-\frac{P(A\cap B)}{P(A)}=\frac{P(B)P(A)-P(A\cap B)P(A\cup B)}{P(A\cup B)P(A)},
$$
which is positive, if and only if $P(B)P(A)\geq P(A\cap B)P(A\cup B)$. Now,
$$
P(A\cap B)P(A\cup B)=P(A\cap B)(P(A\setminus B)+P(B\setminus A)+P(A\cap B))\\
\leq P(A\cap B)(P(A\setminus B)+P(B\setminus A)+P(A\cap B))+P(A\setminus B)P(B\setminus A) \\
=P(A)P(B),
$$
if you write out everything.
Thus, the statement is true.
A: You can approach this considering 3 cases:
case 1: $B \subset A \implies \cfrac{ \mathbb P(B)}{\mathbb P(A \cup B)} =  \cfrac{ \mathbb P(B)}{\mathbb P(A)} \hspace{11mm} \cfrac{ \mathbb P(B \cap A)}{\mathbb P(A)} = \cfrac{ \mathbb P(B)}{\mathbb P(A)}$
Thus, $\hspace{5 mm} \cfrac{ \mathbb P(B)}{\mathbb P(A \cup B)} = \cfrac{ \mathbb P(B \cap A)}{\mathbb P(A)}$
case 2: $A \subseteq B \implies \cfrac{ \mathbb P(B)}{\mathbb P(A \cup B)} =  \cfrac{ \mathbb P(B)}{\mathbb P(B)} = 1 \hspace{11mm} \cfrac{ \mathbb P(B \cap A)}{\mathbb P(A)} = \cfrac{ \mathbb P(A)}{\mathbb P(A)} =1$
Thus, $\hspace{5 mm} \cfrac{ \mathbb P(B)}{\mathbb P(A \cup B)} = \cfrac{ \mathbb P(B \cap A)}{\mathbb P(A)}$
case 3: $A\cap B = \emptyset \implies  \cfrac{ \mathbb P(B \cap A)}{\mathbb P(A)} = \cfrac{ \mathbb P(\emptyset)}{\mathbb P(A)} = 0 \hspace{11 mm}$ Clearly, $ \hspace{5 mm}\cfrac{ \mathbb P(B)}{\mathbb P(A \cup B)}  \geq 0$
case 4: $A\cap B \not = \emptyset \implies \mathbb P (A) > \mathbb P (A \cap B) \hspace {5 mm} \mathbb P (B) > \mathbb P (A \cap B)$
Therefore, $\mathbb P (B) \cdot \mathbb P (A) - \mathbb P(B \cap A) \cdot \mathbb P(B \cup A) = \mathbb P (B) \cdot \mathbb P (A) - \mathbb P(B \cap A) \cdot (\mathbb P(B) + \mathbb P(A) - \mathbb P (B \cap A)) \\ = \mathbb P (B) \cdot \mathbb P (A) - \mathbb P(B \cap A) \cdot \mathbb P(B) - \mathbb P(B \cap A) \cdot \mathbb P(A) + \mathbb P(B \cap A) \cdot \mathbb P(B \cap A) \\ = (\mathbb P(A) - \mathbb P(B \cap A)) \cdot (\mathbb P(B) - \mathbb P(B \cap A)) > 0$
$\mathbb P (B) \cdot \mathbb P (A) - \mathbb P(B \cap A) \cdot \mathbb P(B \cup A) > 0 \implies \cfrac{ \mathbb P(B)}{\mathbb P(B \cup A)} > \cfrac{ \mathbb P(B \cap A)}{\mathbb P(A)}$
A: The inequality boils down to showing that $\mathbb P(A)\mathbb P(B)\geq \mathbb P(A\cap B)\mathbb P(A\cup B)$. The intuition behind why this is true is the same as the intuition behind the AM-GM inequality: the product of two quantities with given sum decreases as the two quantities become more spread out.
The intuition applies to this problem since $\mathbb P(A\cap B)+\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)$ and $|\mathbb P(A)-\mathbb P(B)|\leq |\mathbb P(A\cup B)-\mathbb P(A\cap B)|$ follows from the more precise identity
$$
|\mathbb P(A\cup B)-\mathbb P(A\cap B)|=|\mathbb P(A)-\mathbb P(B)|+2\min\bigl(\mathbb P(A\setminus B),\mathbb P(B\setminus A)\bigr),
$$
which can be seen for example by drawing a Venn diagram. Consequently
$$
\bigl(\mathbb P(A)+\mathbb P(B)\bigr)^2-\bigl(\mathbb P(A)-\mathbb P(B)\bigr)^2\geq\bigl(\mathbb P(A\cap B)+\mathbb P(A\cup B)\bigr)^2-\bigl(\mathbb P(A\cup B)-\mathbb P(A\cap B)\bigr)^2,
$$
which (after expanding squares and cancelling a factor of $4$) yields the desired inequality.
