$A=\{(x_1,x_2,x_3): x_1\le x_2 \le x_3 \}, B=\{(x_1,x_2,x_3): x_2\le x_1 \le x_3 \}$. Show $P(Z+\mu \in A) \ge P(Z+ \mu \in B)$ for $\mu \in A$. Let  $A=\{(x_1,x_2,x_3):  x_1\le x_2 \le x_3 \}$ and $B=\{(x_1,x_2,x_3):  x_2\le x_1 \le x_3 \}$. 
Let $Z \in \mathbb{R}^3$ be a standard normal random vector.    
I am trying to see if the following inequality holds:
\begin{align}
P(Z+\mu \in  A) \ge  P(Z +\mu\in B),
\end{align} 
for all $\mu \in A$. 
Here is the proof that I tried that doesn't work
\begin{align}
P(Z+\mu \in  A)& \ge  P(Z \in  A)\\
&= P(Z \in  B)\\
&\ge P(Z+\mu \in  B).
\end{align} 
The first inequality and equality are true.  However, the last inequality does not hold as was shown here.
Nonetheless, I think the inequality is true. This is also related to another question I asked here. 
 A: The claim is true, but the shapes of $A$ and $B$ are just weird enough that I can't find a way to prove this by transforming $\mu$ or $Z$.  Instead I had to transform the destination.
Let $\mu = (c_1, c_2, c_3) \in A$, and consider some $b = (b_1, b_2, b_3) \in B$.  Its "transformed partner" is $a = T(b) = (b_2, b_1, b_3) \in A$.  The key is:
Lemma: $|\mu - a| \le |\mu - b|$.  
Proof:
$$|\mu - a|^2 = (c_1 - b_2)^2 + (c_2 - b_1)^2 + (c_3 - b_3)^2$$
$$|\mu - b|^2 = (c_1 - b_1)^2 + (c_2 - b_2)^2 + (c_3 - b_3)^2$$
$$|\mu - a|^2 - |\mu - b|^2 = -2c_1 b_2 -2c_2 b_1 + 2 c_1 b_1 + 2 c_2 b_2 = -2(c_2 - c_1)(b_1 - b_2) \le 0$$
where the last inequality comes from $c_2 \ge c_1 (\mu \in A), b_1 \ge b_2 (b \in B).  \square$
Now since the $T()$ mapping is bijective, measure-preserving, rigid, etc., the claim follows:
$$P(Z + \mu \in B) = \iiint_B \phi(|\mu - b|) \,db \le \iiint_A \phi(|\mu - a|) \,da = P(Z + \mu \in A)$$
where $\phi()$ is the pdf for vector $Z$, and I have written it in a form to highlight the fact that it only depends on $|Z|$.
