Why does $\|x\|_2 \le \|x\|_1$?

Question

Why does this inequality always hold true? $$\|x\|_2 \leq\|x\|_1$$ where $\|x\|_2$ is the euclidean norm : $\|x\|_2 = \sqrt{x_1^2 + \dots + x_n^2}$ , and $\|x\|_1$ the $L_1$ norm : $\|x\|_1 = |x_1| + \dots + |x_n|$.

Answer 1

Hint: Compare $\|x\|_2 ^2$ and $\|x\|_1 ^2$.

Answer 2

$||x||_2^2=x_1^2+...+x_n^2\leq x_1^2+...x_n^2+2\sum_{i<j}|x_i||x_j|=(|x_1|+...+|x_n|)^2=||x||_1^2$

Answer 3

Note that $B_1(0,1) = \{ x | \|x\|_1 \le 1 \} \subset B_2(0,1) =\{ x | \|x\| \le 1 \} $.

Note that $\|x\|_p = \inf \{ t > 0 | x \in t B_p(0,1)\}$, $p=1,2$ (Minkowski functional).

Since $\{ t > 0 | x \in t B_1(0,1)\} \subset \{ t > 0 | x \in t B_2(0,1)\}$ we must have $\|x\|_2 \le \|x\|_1$, since we are taking the $\inf$.