Show that $ P(A \cap B \mid A \cup B) \le P(A \cap B \mid A) $, where $A$ and $B$ are events and $P(A) > 0$ 
Show that if $A$ and $B$ are events and $P(A) > 0$, then $P(A \cap B \mid A \cup B) \le P(A \cap B \mid A)$. 

I have proceeded as follows:
\begin{align}
& P(A \cap B \mid A \cup B)
\le P(A \cap B \mid A) \\
\iff & \frac{P((A \cap B) \cap (A \cup B))}{P(A \cup B)}
\le \frac{P((A \cap B) \cap A)}{P(A)} \\
\iff & \frac{P([(A \cap B) \cap A] \cup [(A \cap B) \cap B])}{P(A \cup B)} \le \frac{P(A \cap B)}{P(A)} \\
\iff & \frac{P((A \cap B) \cup (A \cap B)}{P(A \cup B)}
\le \frac{P(A \cap B)}{P(A)} \\
\iff & \frac{P(A \cap B)}{P(A \cup B)}
\le \frac{P(A \cap B)}{P(A)} \\
\iff & \frac{1}{P(A) + P(B) - P(A \cap B)}
\le \frac{1}{P(A)} \\
\iff & P(A)
\le P(A) + P(B) - P(A \cap B) \\
\iff & 0 \le P(B) - P(A \cap B) 
\end{align}
which is obviously true.
Also, I am dividing by $P(A \cap B)$ in the $5^{th}$ implication, since in the case that $P(A \cap B) = 0$ the inequality trivially holds true.
I just wanted to verify the correctness of my work, and if, perhaps, there is a more succinct demonstration.
 A: A shortcut:
No need to distribute the intersection with respect to union. Just say $A\cap B\subseteq A\cup B$, so $(A\cap B)\cap (A\cup B)=A \cap B$
Second shortcut:
There is no need to expand $P(A\cup B)$, just say that $A\subseteq A\cup B$, so $P(A) \le P(A\cup B)$, and $\frac{1}{P(A\cup B)}\le \frac{1}{P(A)}$
Also, before you divide with $P(A\cup B)$, it might be good to say what happens when $P(A\cup B)=0$ separately
A: The direction in which you are deriving is a bit problematic. Using the mere definition can give you the more formal and simple proof:
$$\Pr \left( A\cap B|A\cup B\right) =\dfrac {\Pr \left( A\cap B\cap \left( A\cup B\right) \right) }{\Pr \left( A\cup B\right) }=\dfrac {\Pr \left( A\cap B\right) }{\Pr \left( A\cup B\right)}$$
Using, again the fact that 
$$C\subseteq D\Rightarrow \Pr \left( C\right) \leq \Pr \left( D\right) $$
And we get that the above right hand side
$$\leq \dfrac {\Pr \left( A\cap B\right) }{\Pr \left( A\right) }=\dfrac {\Pr \left( A\cap B\cap A\right) }{\Pr \left( A\right) }=\Pr \left( A\cap B|A\right).$$
A: If $P(A \cap B) = 0$, both sides are $0$ and the inequality holds, so assume the contrary. Note the following:
(i) $(A \cup B) \cap (A \cap B) = A\cap B$
(ii) $(A \cap B) \cap A = A \cap B$
(iii) $A \cap B \subseteq A \implies P(A \cap B) \le P(A)$
The first two are consequences of the fact that $C \subseteq D \implies C \cap D = C$ (or you can just draw a Venn Diagram). The third is a basic property of probability measures.
So:
$P(A \cap B  | A \cup B) = \cfrac{P((A \cap B) \cap (A \cup B))}{P(A \cup B)} = \cfrac{P(A \cap B)}{P(A \cup B)}$,
where (i) was used in the second equality, and
$P(A \cap B  | A) = \cfrac{P((A \cap B)  \cap A)}{P(A)} = \cfrac{P(A \cap B)}{P(A)}$,
where (ii) was used in the second equality.
Inverting the inequality in (iii) and multiplying by $P(A \cap B)$ on both sides gives the result.
Your proof is correct, by the way.
