Is E[|X+Y|]\leq E[|X| ] + E[|Y|] Let $X , Y$ be two independent Random variables then
is 
$E[|aX+bY|] \leq |a|E[|X|]+|b|[E[|Y|]$ ?
where $a ,b \in \mathbb{R}$.
Thank you.
 A: Yes, no independence needed: $$\mathbb{E} \| aX + bY \| = \int \| aX + bY \| \leqslant \int \|aX\| + \|bY\| $$$$ = \|a\| \int \|X\| + \|b\| \int \|Y\| = \|a\|\mathbb{E}\|X\| + \|b\|\mathbb{E}\|Y\|$$where I used the triangle inequality and the monotonicity + linearity of integration. 
A: Similar solution to @uncookedfalcon's answer, but with different notation would be as follows:
\begin{align}
|aX+bY| &\leq |aX| + |bY|  & \text{triangle inequality} \\
|aX+bY| &\leq |a|\cdot|X| + |b|\cdot|Y|  & \text{norm homogeneity} \\
\mathbb{E}\Big[|aX+bY|\Big] &\leq \mathbb{E}\Big[|a|\cdot|X| + |b|\cdot|Y|\Big]  & \text{expected value monotonicity} \\
\mathbb{E}\Big[|aX+bY|\Big] &\leq |a|\cdot\mathbb{E}\Big[|X|\Big] + |b|\cdot\mathbb{E}\Big[|Y|\Big]  & \text{expected value linearity} \\
\end{align}
In fact this exactly the same with $\mathbb{E}[X] = \int_\Omega X\ \mathrm{d}P$, but is nicer if you don't want to mention the integral symbol while $\mathbb{E}[X] = \sum_{k \in \mathrm{Cod}(k)}k\cdot P(X = k)$ or $\mathbb{E}[X] = \sum_{\omega \in \Omega}X(\omega) \cdot P(\omega)$. I know all this is just a difference in notation (i.e. the sum is still the integral for appropriate measure), but sometimes it matters.
I hope this helps ;-)
