Show that $||x||_2\leq ||x||_1$ on $\Bbb R^d$

Question

My attempt:

First I will prove that $\sum^n_{i=1}x^2_i\leq \sum^n_{i=1}|x_i|$:

Let $x=(x_1,x_2)\in \Bbb R^2$ Then $(x_1+x_2)^2=x^2_1+x^2_2+2x_1x_2\geq x_1^2+x^2_2$

Hence $\sqrt{x_1^2+x_2^2}\leq |x_1|+|x_2|$ . So by induction, $\sum^n_{i=1}x^2_i\leq \sum^n_{i=1}|x_i|$

The above proof I took from this answer

So, $||x||_2=\sqrt{\sum^d_{i=1}x^2_i}\leq \sum^d_{i=1}|x_i|=||x||_1$

Would this be correct?

The inequality in the fourth line is incorrect. Counterexample: take $x_1=1$ and $x_2=-1$. Then $(x_1+x_2)^2 = 0$, but $x_1^2 + x_2^2 = 2$. — Ernie060, May 11 '20 at 15:19
If you are familiar with metric space and topology you may have a look at this: any norm on $\mathbb R^d$ are equivalent — Noob mathematician, May 11 '20 at 15:44
Your argument would be correct if instead of $(x_1+x_2)^2 = x_1^2+x_2^2+2x_1x_2 \geq x_1^2+x_2^2$ you wrote $(|x_1|+|x_2|)^2 = x_1^2+x_2^2+2|x_1x_2| \geq x_1^2+x_2^2$. — Jakobian, Sep 07 '21 at 16:44

score 2 · Accepted Answer · answered May 11 '20 at 15:42

Let $x= (x_{1},...,x_{n})\in \mathbb{R}^{n}$. Let us define: $$ x^{(1)} = (x_{1},0,...,0)$$ $$x^{(2)} = (0,x_{2},...,0) $$ $$\vdots $$ $$x^{(n)} = (0,0,...,x_{n})$$ Then, we have: $$x = x^{(1)}+\cdots+x^{(n)}$$ We can use the triangle inequality for norms to get: $$||x||_{2} = ||x^{(1)}+\cdots+x^{(n)}||_{2} \le ||x^{(1)}||_{2}+\cdots+||x^{(n)}||_{2} = \sqrt{x_{1}^{2}}+\cdots+\sqrt{x_{n}^{2}} = |x_{1}|+\cdots+|x_{n}| = ||x||_{1}$$

Show that $||x||_2\leq ||x||_1$ on $\Bbb R^d$

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