How can I establish an inequality between $|\frac1n \sum_{i=1}^nX_i|$ and $\frac1n\sum^n_{i=1}|X_i|$ where $X_i \sim N(0,1)$? 
Let $X_1, \ldots , X_n$ be a random sample from a $N(0,1)$ population. Define $Y_1=|\frac1n \sum_{i=1}^nX_i|$ and $Y_2=\frac1n\sum^n_{i=1}|X_i|$. Find a relationship between $E(Y_1)$ and $E(Y_2)$.

I have a feeling that I will need to use Jensen's Inequality here. Since $X_i \sim N(0,1)$, the linear combination $\sum_{i=1}^nX_i \sim N(0,n)$.
$E(Y_1)=\frac1n \cdot E(|\sum_{i=1}^nX_i|)$
$E(Y_2)=\frac1n \cdot E(\sum_{i=1}^n|X_i|)$
However, I am not too sure how to compute $E(Y_1)$ or $E(Y_2)$ from this step since there are absolute values. Do I have to use a cdf approach?
 A: Assuming independent $X_i$, the mean $\frac{1}{n}\sum X_i$ is also normal, i.e. $N(0,1/n)$. Absolute value of it is Half-normal, which has mean $E[Y_1]=\frac{\sigma\sqrt{2}}{\sqrt{\pi}}=\sqrt{\frac{2}{n\pi}}$. For $Y_2$ we can find the expected value directly:
$$E[Y_2]=\frac{1}{n}\sum_{i=1}^n E[|X_i|]=E[|X_i|]=\sqrt\frac{2}{\pi}$$
This means $\sqrt n E[Y_1]=E[Y_2]$. I think an equality is better than inequality.
A: Answer: 
Whatever the distribution of $X_1,...,X_n$,
$$\mathbb{E} Y_2 \geq \mathbb{E} Y_1.$$
Details:
For any $n$ numbers $X_1,..., X_n$ it is true that 
$$ \sum_i |X_i| \geq |\sum_i X_i|$$
and dividing both sides by $n$:
$$ \frac{1}{n}\sum_i |X_i| \geq \frac{1}{n}|\sum_i X_i| = |\frac{1}{n}\sum_i X_i|.$$
Now, the key word is 'any', that means that if we take integrals of both sides of the above inequality, the integrand of the left hand side is uniformly higher than the integrand of the right hand side, i.e. 
$$ \mathbb{E} Y_2 = \int \frac{1}{n}\sum_i |X_i| dX_1 ...dX_n \geq \int |\frac{1}{n}\sum_i X_i| dX_1...dX_n = \mathbb{E} Y_1.$$
