This is problem 1-1 from Spivak Calculus on Manifolds
My proof of $|x| \leq \sum |x_i|$ is slightly different from others I've seen. I would like to know if the proof is correct, but also -if it is correct- if it is not quite as good as the standard one. (less rigoruous?) (This is the usual proof: problem 1-1 http://vision.caltech.edu/~kchalupk/spivak.html
$|x| \equiv \sqrt{{\sum(x_i)^2}}$
For clarity, take $i=2$.
$|x||x|=|x_1|^2+|x_2|^2$
$|x|=|x_1|\frac{|x_1|}{|x|}+|x_2|\frac{|x_2|}{|x|}$
$\frac{|x_i|}{|x|} \leq 1$
$|x| \leq |x_1|+|x_2|$
For the general case take $i=n$.
The idea is basically correct. Of course you have to assume $x\ne0$ (but in this case the inequality trivially holds).
First you note that, for $x=(x_1,\dots,x_n)\ne0$, $$ |x_i|\le|x|=\sqrt{|x_1|^2+\dots+|x_n|^2} $$ just by squaring. Then you have $$ 0\le \frac{|x_i|}{|x|}\le 1 $$ so $$ |x|=\sum_{i=1}^n|x_i|\frac{|x_i|}{|x|}\le\sum_{i=1}^n|x_i| $$