Spivak: Calculus on Manifolds Norms (Problem 1.1) I found it difficult to prove this problem due to the following fact. I want to derive it myself but am having trouble laying out the proof. First, I want to show the following holds:
Show that $$(\sum_{i=1}^n |x_i|)^2= \sum_{i=1}^n x_{i}^2 + 2\sum_{i\neq j}|x_i||x_j|$$
I would also like a little more emphasis on the notation being used for the second sum. Thanks!
 A: An helpful way would be to arrange the terms $|x_1|,...,|x_n|$ in a square array in an increasing order. Regards this square array as a matrix of size $n\times n$ whose $ij$ entry is $|x_i||x_j|$ for $1\leqslant i,j\leqslant n$. The diagonal entries are $|x_i||x_i|=|x_i|^2$ and the off-diagonal entries are given by $|x_i||x_j|$ for $i\neq j$. The total sum of all entries would be 
$$\sum_{i,j}|x_i||x_j|=\sum_{i=1}^n|x_i|^2+\sum_{i>j}|x_i||x_j|+\sum_{i<j}|x_i||x_j|=\sum_{i=1}^n|x_i|^2+2\sum_{i\neq j}|x_i||x_j|$$
where we used $|x_i||x_j|=|x_j||x_i|$ in the very last step (i.e. the matrix is symmetric). This total is also equal to 
$$\Big(\sum_{i=1}^n|x_i|\Big)^2$$
A: Here is a hint. Note that we can expand the first sum as:
$$
\left(\sum_{i=1}^n\lvert x_i\rvert\right)^2 = (\lvert x_1\rvert + \lvert x_2\rvert + \cdots + \lvert x_n\rvert)^2
$$
Expanding this out we have:
$$
(\lvert x_1\rvert + \lvert x_2\rvert + \cdots + \lvert x_n\rvert)^2 = \lvert x_1\rvert\lvert x_1\rvert +\lvert x_1\rvert\lvert x_2\rvert +\lvert x_1\rvert\lvert x_3\rvert +\cdots+\lvert x_1\rvert \lvert x_n\rvert + \lvert x_2\rvert\lvert x_1\rvert +\cdots +\lvert x_n\rvert\lvert x_n\rvert
$$
Grouping together the like terms $x_1x_2=x_2x_1$ we note that we will always pick up two of each pair $i\neq j$, and for each time we have $i=j$ we get a square term. Then of course:
$$
\left(\sum_{i=1}^n\lvert x_i\rvert\right)^2 = \sum_{i=1}^n x_i^2 + 2\sum_{i\neq j} \lvert x_i\rvert\lvert x_j\rvert
$$
See if you can use this hint to make a more rigorous argument for why this is true. 
A: The result gives you the specific quadratic sums we all learnt at high school:
$
(a+b)^2 = a^2 + b^2 + 2ab; 
$
$
(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca; 
$
and in general
$
(a_1 + a_2 ... a_n)^2 = a_1^2 + a_2^2.... a_n^2 + 2a_1a_2 + 2a_2a_1+ 2a_1a_3 + 2a_3a_1 +... pair\ wise\ terms  
$
Therefore
$
(\sum_i{|x_i|})^2 
$
can just be expanded as
$
 |x_1|^2 + |x_2|^2 .... |x_n|^2 + 2*\sum_{i\neq j}{|x_i|*|x_j|}
$
which gives you the result you seek
I actually find it weird to have the $||$ notation being somewhat abused in the chapter. Because $|x|$ is the norm of a vector, ${x}$, and $|\int_a^b{f^2}|$ (Example 1-6) is just its absolute value. 
The $|x|$ in that equation seem necessary for the sign to be right for something like $(a-b+c)^2$ but I would much rather prefer if this was corrected by using the modern $||x||$ for the norm and reserve the other for absolute value.
