Proof about vector mean in the euclidean norm I am trying to show that, given a sequence of real vectors $v^1,\dots,v^n$, the following is true:
$$\left\lVert \frac{1}{n}\sum_{i=1}^n v^i \right\rVert_2^2 \leq \frac{1}{n} \sum_{i=1}^n\left\lVert v^i \right\rVert_2^2.$$
I am not a 100 % sure that it's true, but I haven't been able to find any counter example.
Thank you.
 A: Let $W = \mathbb{R^n}$ with the standard inner product where for $x = (x_1,\cdots, x_n),$ $y = (y_1, \cdots, y_n) \in W$ $\langle x,y \rangle = x_1y_1 + \cdots + x_n y_n$.
Let $x = (1, 1, \ldots, 1, 1),$ $y = (\lVert v_1 \rVert, \lVert v_2 \rVert, \ldots, \lVert v_{n-1} \rVert, \lVert v_n \rVert) \in W$ 
By the Cauchy-Schwarz inequality
$$\vert\langle x, y\rangle\vert^2 \le \langle x,x\rangle\cdot\langle y,y\rangle$$
Thus,  
$\begin{align}
\vert1 \cdot \lVert v_1 \rVert + 1 \cdot \lVert v_2 \rVert + \ldots + 1 \cdot \lVert v_n \rVert\vert^2 &\le (1\cdot1 +\ldots+1\cdot1)(\lVert v_1 \rVert\cdot\lVert v_1 \rVert + \ldots + \lVert v_n \rVert\cdot\lVert v_n \rVert)\\ 
\vert\lVert v_1 \rVert+ \ldots +\lVert v_n \rVert\vert^2 &\le n(\lVert v_1 \rVert^2 + \ldots + \lVert v_n \rVert^2)\\
\text{By the Triangle Inequality we get,}\\
\lVert v_1 + \ldots + v_n\rVert^2 \le \vert\lVert v_1 \rVert+ \ldots +\lVert v_n \rVert\vert^2 &\le n(\lVert v_1 \rVert^2 + \ldots + \lVert v_n \rVert^2)\\
\lVert v_1 + \ldots + v_n\rVert^2 &\le  n(\lVert v_1 \rVert^2 + \ldots + \lVert v_n \rVert^2)\\
{1\over n^2} \lVert v_1 + \ldots + v_n\rVert^2 &\le {1 \over n}(\lVert v_1 \rVert^2 + \ldots + \lVert v_n \rVert^2)\\
\lVert { 1 \over n} (v_1 + \ldots + v_n)\rVert^2 &\le {1 \over n}(\lVert v_1 \rVert^2 + \ldots + \lVert v_n \rVert^2)\\
\left\lVert \frac{1}{n}\sum_{i=1}^n v_i \right\rVert^2 &\leq \frac{1}{n} \sum_{i=1}^n\left\lVert v_i \right\rVert^2.
\end{align}$
This is nice problem because it involves the Cauchy-Schwarz inequality and the Triangle Inequality, two characteristic inequalities of inner product spaces.  Please let me know if you have any questions.
