Prove if $P(A \cup B) \le P(A \cap B)$, then $P(A) = P(B)$. I am unsure if my thought process for proving this inequality is correct. This is what I have so far:
$$
P(A \cup B) = P(A) + P(B) - P(A \cap B) ≤ P(A \cap B)
$$
$$
P(A) + P(B) \le 2  P(A \cap B)
$$
$$
P(A) + P(B) \le 2 P(A) P(B)
$$
and in order for the last line to be possible, $P(A)$ and $P(B)$ have to be $0$, thus $P(A) = P(B)$. This doesn't seem right because I feel like there should be more cases than just $0$ for this inequality to be true and I was wondering if anyone could help guide me in the right direction.
 A: Edit: See Amit's answer instead for a quicker proof using the same idea.

Hint. We always have $P(A \cap B) \le P(A \cup B)$ since $A \cap B \subseteq A \cup B$.
Thus, your hypothesis tells us that $P(A \cup B) = P(A \cap B)$. As you have noted, inclusion-exclusion gives us $$P(A \cup B) = P(A) + P(B) - P(A \cap B).$$
Thus, we get $$2P(A \cup B) = P(A) + P(B).$$
We can rewrite this as $$[P(A \cup B) - P(A)] + [P(A \cup B) - P(B)] = 0.$$
Do you see that both the $[\cdots]$ terms are non-negative? What can we say now?
A: Since $A \cap B \subseteq A \subseteq A \cup B$, we have $P(A\cap B) \leq P(A) \leq P(A\cup B)$. The two extreme quantities are equal (Why?), so $P(A)$ must be equal to these too. A similar argument works out for $P(B)$.
A: In fact, no exclusion-inclusion formula is needed. Just use
\begin{align*}
& P(A) \leq P(A \cup B) \leq P(A \cap B) \leq P(B), \\
& P(B) \leq P(A \cup B) \leq P(A \cap B) \leq P(A)
\end{align*}
gives you $P(A) = P(B)$.
A: Editing added to the very end of my answer, where I disagree with the comment of Zhanxiong, which follows my answer.

Starting with your first equation, you also have that 
$p(A) \geq p(A \cap B)$ and $p(B) \geq p(A \cap B)$. 
Therefore, $p(A) + p(B) \geq 2 p(A\cap B).$
Therefore, you can now conclude that $p(A) + p(B) = 2p(A \cap B)$.
The only way that this is possible is if 
$p(A) = p(A\cap B)$ and $p(B) = p(A\cap B)$. 
Edit
Thanks to Zhanxiong (re comment below) for catching my mistaken conclusion that the above implies that $A \subseteq B$ and that $B\subseteq A.$
Fortunately, that mistaken conclusion is unnecessary to complete the analysis.
Since $p(A) = p(A \cap B) = p(B)$ 
you have that $p(A) = p(B).$

Edit
On second thought, I disagree with the comment of Zhanxiong, following my answer.
That is, clearly $p(A) = p(A\cap B) + p(A \cap [B^c])$, 
where $B^c$ represents the complement of $B$.  Thus, if $p(A) = p(A\cap B)$, then $p(A \cap [B^c])$ must $ = 0.$ 
This implies that $A$ and $B^c$ are disjoint, which implies that $A \subseteq B.$
Rebuttal from Zhanxiong 
A basic fact in probability, $P(E)=0$ does not imply $E$ is an empty set.
A: A slightly different approach from the others: we have the identity
$$A\cup B = (A\cap B) \cup (A\triangle B)$$
where $A\triangle B = (A\setminus B)\cup(B\setminus A)$ - this is called the symmetric difference of the sets. Draw it on a Venn diagram to see why it has this name if you're unsure. One very nice aspect here is that $A\cap B$ and $A\triangle B$ are actually mutually distinct, i.e. their intersection is empty. This is also true for the two individual pieces in the symmetric difference, i.e. $A\setminus B$ and $B\setminus A$. This allows you then to say that
$$ P(A\cup B) = P((A\cap B)\cup(A\triangle B)) = P(A\cap B) + P(A\triangle B).$$
We have not yet invoked your inequality. Let's do that now. From your inequality, we have that
$$ P(A\cup B) \le P(A\cap B) \Longrightarrow P(A\cap B) + P(A\triangle B) \le P(A\cap B). $$
Notice the appearance of $P(A\cap B)$ on both sides. Eliminating it gives us that
$$ P(A\triangle B) = 0. $$
This then tells us that $P(A\cup B) = P(A\cap B)$, however $P(A) \le P(A\cup B)$ and $P(A\cap B)\le P(A)$, so $P(A) \le P(A\cup B) = P(A\cap B) \le P(A)$ so that $P(A) = P(A\cup B) = P(A\cap B)$. Similar for $B$ which gives you equality between $P(A)$ and $P(B)$.
