Showing the Metric Function Satisfies the Triangle Inequality Let $d(X,Y)=E\left[ \frac{|X-Y|}{1+|X-Y|}\right]$. How can we show
$$d(X,Z)\leq d(X,Y)+d(Y,Z)?$$
 A: You basically only need two things here: 


*

*the first as Eric points out is the linearity of the expectation

*the second is the easy to verify inequality $\frac{\lvert a+b\rvert}{1+\lvert a+b\rvert}\leq \frac{ \lvert a\rvert+\lvert b\rvert}{1+\lvert a\rvert+\lvert b\rvert} $ (you can verify this by cross multiplying and using the fact, that the absolute value is a norm)
After this, it is essentially just plugging in. We get for real valued random variables $X,Y,Z$:
$$
d(X,Z)=E\left[ \frac{\lvert X-Z+Y-Y\rvert}{1+\lvert X-Z+Y-Y\rvert} \right]=E\left[ \frac{\lvert X-Y+Y-Z\rvert}{1+\lvert X-Y+Y-Z\rvert} \right]
$$
Now use the facts from the above two points and that $\lvert\cdot\rvert\geq0$ (important for the upcoming denomintor), we eventually get 
\begin{align}
E\left[ \frac{\lvert X-Y+Y-Z\rvert}{1+\lvert X-Y+Y-Z\rvert} \right]&\leq E\left[ \frac{\lvert X-Y\rvert}{1+\lvert X-Y\rvert+\lvert Y-Z\rvert} \right]+E\left[ \frac{\lvert Y-Z\rvert}{1+\lvert X-Y\rvert+\lvert Y-Z\rvert} \right]\\
&\leq E\left[ \frac{\lvert X-Y\rvert}{1+\lvert X-Y\rvert} \right]+E\left[ \frac{\lvert Y-Z\rvert}{1+\lvert Y-Z\rvert} \right]=d(X,Y)+d(Y,Z)
\end{align}
