The expectation of absolute value of random variables

I need some help with the following problem:

Let $X_1,...,X_n$ be a random sample from Normal$(0,1)$ population. Define $$Y_1=| {{1 \over n}\sum_{i=1}^{n}X_i}|, \ Y_2={1 \over n}\sum_{i=1}^{n}|X_i|.$$ Calculate $E[Y_1]$ and $E[Y_2]$, and establish the inequality between them.

I may feel this should not be a very hard problem but I did get stuck somewhere. And I know it is $E[Y_1]\le E[Y_2]$ and I could prove this. But can anyone help me with how to exact find $E[Y_1]$ and $E[Y_2]$?

Thanks in advance.

It is worth knowing that the expected absolute value of a normal random variable with mean $\mu = 0$ and standard deviation $\sigma$ is $\sigma \sqrt{\dfrac{2}{\pi}}$. See Wikipedia on the half-normal distribution.

$Y_1$ is the absolute value of a normal random variable with mean $0$ and standard deviation ${\dfrac{1}{\sqrt{n}}}$ so $E[Y_1] = \sqrt{\dfrac{2}{\pi n}}$.

$Y_2$ is the average of $n$ absolute values of normal random variables with mean $0$ and standard deviation $1$ so the average of random variables with expected value $\sqrt{\dfrac{2}{\pi}}$ meaning $E[Y_2] = \sqrt{\dfrac{2}{\pi}}$.

For $n \gt 1$ you have $\sqrt{\dfrac{2}{\pi n}} \lt \sqrt{\dfrac{2}{\pi }}$.

Regarding the general form of the mean of the absolute value of a Normal$(\mu,\sigma^2)$:

Following @Henry's link to Wikipedia's article on the half-normal, I found that there is now a page for the so-called "Folded normal distribution", which is an extension of the half-normal for $\mu\neq 0$. They give the expression for the mean, which coincides with the one given by @Henry for $\mu=0$

$$ \sigma \sqrt{\frac{2}{\pi}} e^{-\frac{\mu^2}{2\sigma^2}} + \mu \left(1 - 2\Phi\left(\frac{-\mu}{\sigma}\right) \right) $$