How to show $|E(X)| \leq E(|X|)$ for column vector $X$ of random variables Suppose $X=(X_1,X_2,...,X_d)^T$ is a column vector of random variables and $| \cdot |$ is the euclidean norm. Expectations are taken component-wise. Then how do we show:
$$|E(X)| \leq E(|X|) \quad ?$$
Many thanks for your help. 
EDIT:
Following the hints below I have started by letting $X = \sum_{k=1}^d x_k 1_{A_k}$
where $x_k = (x_k^1,...,x_k^d) \in \mathbb{R}^d$ and where $A_k$ are disjoint measurable sets in the underlying probability space. By approximation this is sufficient.
The claim to prove is now:
$$\left( \sum_i \left( \sum_k x^i_k P(A_k) \right)^2 \right)^{1/2} \leq E \left( \left( \sum_{i,k} (x^i_k)^2 1_{A_k} \right)^{1/2} \right) .$$
By Jensen's inequality,
$$ RHS \geq  \sum_k P(A_k)  \left( \sum_i (x_k^i)^2\right)^{1/2}$$
and also
$$ LHS \leq \left( \sum_k P(A_k) \sum_i (x^i_k)^2 \right)^{1/2} .$$
But now Jensen is no good to conclude, as it's "the wrong way round" to take the square root inside in the estimate of the LHS. 
Many thanks for your suggestions. 
 A: By the triangle inequality and positive homogeneity, the Euclidean-norm, $\|\ \|:C\to\mathbb R$, is convex (where $C\subseteq \mathbb R^d$ is a convex set). That is, for $\lambda\in[0,1]$ and $x,y\in C$,
$$
\|\lambda x + (1-\lambda)y\|\leqslant \lambda \|x\|+(1-\lambda)\|y\|\ .
$$
As a consequence, there exists a linear real-valued map, $l:C\to\mathbb R$, such that 


*

*$l(x)\leqslant\|x\|$ for all $x\in C$; 

*$l(\mathbb E[X])=\|\mathbb E[X]\|$ for integrable $d$-dimensional random vector $X$.   


Therefore,
$$
\| \mathbb E[X] \| = l(\mathbb E[X]) = (\mathbb E[l(X)]) \leqslant \mathbb E[\| X \|]\ .
$$
As hinted in the comments, this is simply an application of Jensen's inequality.
A: Remember that for $x\in \mathbb R^d$, $\|x\|=\sup\{|f(x)|: f\in (\mathbb R^d)^* \text{ and } \|f\|_*\leq 1\}$ where $(\mathbb R^d)^*$ is the topological dual and $\|\cdot\|_*$ is the dual norm.
Note also that if $x\in \mathbb R^d$ and $f\in (\mathbb R^d)^*$,
$f(x) = \sum_{i=1}^d x_i f(e_i)$.
Thus $\|E[X]\|= \sup\{|f(E[X])|: f\in (\mathbb R^d)^* \text{ and } \|f\|_*\leq 1\}$.
Next, fix some $f\in (\mathbb R^d)^*$ with $\|f\|_*\leq 1$ and note that
$|f(E[X])| = |\sum_{i=1}^d E[X_i] f(e_i)|=|E[\sum_{i=1}^d X_i f(e_i)]| = |E[f(X)]|\leq E[|f(X)|]\leq E[\|X\|]$ where the last inequality follows from
$\|f\|_*\leq 1$.
Taking the supremum yields the claim.
