How to solve it?
Let $X_1,X_2,\dots,X_n$ be a random sample from a $N(0,1)$ population. Define $$ Y_1=\left|\frac{1}{n}\sum_{i=1}^nX_i \right|,\ \ Y_2=\frac{1}{n}\sum_{i=1}^n|X_i|. $$ Calculate $E[Y_1]$ and $E[Y_2]$ and establish the inequality between them.
According to me $$E[Y_1]=E\left[ \left|\frac{1}{n}\sum_{i=1}^nX_i \right|\right]=\frac{1}{n}\left|E\sum_{i=1}^nX_i\right|$$ and $$E[Y_2]=\frac{1}{n}\sum_{i=1}^nE[|X_i|]$$
If $X$ is normally distributed with mean $0$ and variance $1$, then $$ \mathbb{E}[|X|]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}|x|e^{-\frac{x^2}{2}}\;dx=\sqrt{\frac{2}{\pi}}\int_0^{\infty}xe^{-\frac{x^2}{2}}\;dx=\sqrt{\frac{2}{\pi}}$$ and hence $$ \mathbb{E}[Y_2]=\frac{1}{n}\sum_{i=1}^{n}\mathbb{E}[|X_i|]=\sqrt{\frac{2}{\pi}} $$
For $Y_1$ you cannot bring the expectation inside the absolute value. However, observe that $\frac{1}{n}\sum_{i=1}^nX_i$ is normally distributed with mean $0$ and variance $\frac{1}{n}$, so a calculation very similar to the variance $1$ case above will allow you to compute $\mathbb{E}[Y_1]$.