Proof that $E(|X_1 - X_2|)$ is bound by twice the mean Let $X_1$, $X_2$ be iid random variables.
How do I show that for non-negative variables $E(|X_1 - X_2|)$ is bound from above by twice the expected value of $X_1$ (or $X_2$)?
 A: as pointed out in the comments by @zhanxiong, the triangle inequality is sufficient here, take:
$|X_1 -X_2| \leq |X_1| +|X_2|$ and take expectations to get
$\mathbb{E}(|X_1 -X_2|) \leq \mathbb{E}(|X_1|) +\mathbb{E}(|X_2|)$. However, you cannot equate the two marginal expectations to be the same value without assuming they have the same marginal distributions. Assuming this you can conclude $\mathbb{E}(|X_1 -X_2|) \leq 2\mathbb{E}(|X_1|)$ and the non-negative assumption allows you to say that $\mathbb{E}(|X_1|) = \mathbb{E}(X_1)$. So that if $\mu=\mathbb{E}(X_1)$ then you have: 
$\mathbb{E}(|X_1 -X_2|) \leq 2\mu$. 
As mentioned by @zhanxiong in the comments this doesn't use independence but does use that the marginals have the same distributions. 
As @whuber points out in a comment this is not possible if the variables could be negative. Here is a simple example showing what happens if they are negative expectations; let $X_i \sim Rademacher(p=1/3)$ where $\Pr(X_i=1)=1/3$ and $\Pr(X_i=-1)=2/3$. The expectation of each is $\mathbb{E}(X_i) =-1/3$. Then with the i.i.d. assumption we can calculate the four probabilities: 


*

*$\Pr(X_1=1, X_2=1)   = 1/9$ and $|1-1| = 0$

*$\Pr(X_1=1, X_2=-1)  = 2/9$ and $|1-(-1)| =2$

*$\Pr(X_1=-1, X_2=1)  = 2/9$ and $|-1-1| =2$

*$\Pr(X_1=-1, X_2=-1) = 4/9$ and $|-1-(-1)|=0$


Multiplying the probabilities of each event (l.h.s.) with the values on the r.h.s. and summing all four numbers gives you $\mathbb{E}(|X_1 -X_2|)=8/9$ and twice the expectation is: $2\mathbb{E}(X_i) = -2/3$. Clearly, $8/9 \nleq -2/3$ so the claim needs additional conditions/assumptions (non-negativity) to be true. 
