Problem with Einstein summation notation Well I've got a bit problem with ESN so let me introduce my problem :
I have in my lesson the demonstration of this propriety without ESN :
If $(x_i)_{1 \le i \le n}$ is an orthogonal familly, then 
$$
\left|\left| \sum_{i=1}^{n} x_i \right|\right|^2 = \sum_{i=1}^{n} ||x_i||^2
$$
Now my problem is to note this two part of equality with ESN :/ 
$$
||x^i||^2= ???
$$
And I also want to know how to note 
$$
\sum_{i=1}^n\left(x_i \mid \sum_{j=1}^n x_j\right)
$$
where $(.|.)$ is dot product :) 
Shadock 
 A: Update:
Sorry if there was any confusion from my initial post.  I've edited this to more fully explain my answer.  If a vector $x$ can be written as $x = (x_1, \ldots, x_k)$ then in Einstein summation notation $x$ can be denoted as $x_i$ if it's covariant or $x^i$ if it's contravariant (I'm not assuming the Lorentz metric for this kind of vector).  The norm of $x$ can then be written as $||x||^2 = x^ix_i$ where the summation is implied by the double index.  To write it out fully, we have 
$$
\underbrace{x^ix_i}_{Not \; components} \;\; =\;\; \sum_{j=1}^n \underbrace{x_j^2}_{components} \;\; =\;\; ||x||^2.
$$
Unfortunately if you have multiple vectors $\textbf{x}^{(1)}, \ldots, \textbf{x}^{(m)}$ where I use the bold-face for emphasis, then I don't think there is a way to express the quantity
$$
\sum_{j=1}^m ||\textbf{x}^{(j)}||^2
$$
In terms of the Einstein summation convention.  As Zhen Lin pointed out in the comments below, the notation convention is used to manipulate the components of a vector, not multiple vectors at one time.  
@Zhen - thanks for your critique.  I would've missed this otherwise.
A: If your vectors are in an euclidean space then you could also write your identity as:
\begin{equation}
(x_i)_a(x_j)_a = \frac{1}{n} \delta_{ij} (x_k)_a (x_k)_a
\end{equation}
Your second expression can't really be put in ENS, there are no repeated indices there. You could consider, as above, that the inner product is eculidean, and hence I think the best you could do for your second expression would be:
\begin{equation}
\sum_{i,j =1}^{n} (x_i)_a (x_j)_a
\end{equation}
