How does one prove that $\|x\|_2\le \|x\|_1$ I intuitively understand that $\|x\|_2 = \sqrt{x_1^2+\dots+x_n^2}\le \sqrt{x_1^2}+\dots+\sqrt{x_n^2}=|x_1|+\dots+|x_n|=\|x\|_1$. But the thing I'm concerned about is how to prove that $\sqrt{x_1^2+\dots+x_n^2}\le \sqrt{x_1^2}+\dots+\sqrt{x_n^2}$?

I find it easier to argue in the following way: the inequality is true when $x=0$, so assume that $x\neq 0$. Then by the homogeneity of the norms we can assume that $||x||_1=1$. Therefore $|x_j|\leq 1$ for all $j$, so 
$$ ||x||_2^2=|x_1|^2+\dots+|x_n|^2\leq |x_1|+\dots+|x_n|=1$$
and taking square roots yields the desired inequality.
