Is the value of $\mathbb E_{\forall i=1,2,\cdots,n,x_i\sim N(0,1)}\,\Vert\vec x\Vert_2=\sqrt{n}$?

Question

I simulated the result with computer and my guess seems to be correct. Assuming that $\mathbb E_{\forall i=1,2,\cdots,n,x_i\sim N(0,1)}\,\Vert\vec x\Vert_2=\sqrt{\mathbb E_{\forall i=1,2,\cdots,n,x_i\sim N(0,1)}\,\sum_{i=1}^nx_i^2}$, we can easily deduce that the value is $\sqrt n$. However, I am not sure whether the assumption is correct, e.g. a well-known case that $(\mathbb E\,\mathrm x)^2\neq\mathbb E[\mathrm x^2]$ (the difference of the two being the variance of $\mathrm x$).

Answer 1

This equality cannot hold. C-S inequality gives $E\|X\| \leq \sqrt {E \sum X_i^{2}}=\sqrt n$ and equality can hold only when $\|X\|$ is constant. In this this is not a constant so we have strict inequality.