Use the Cauchy-Schwarz inequality to prove $||x||_1 \leq \sqrt{n}||x||_2$

Question

If $x,y \in \mathbb{R}^n$, then the Cauchy-Schwarz Inequality tell us that $$|x^Ty| \leq ||x||_2||y||_2,$$ where \begin{align*} \|x\|_2 &= \sqrt{\sum_{i=1}^nx_i^2},\\ \|x\|_1 &= \sum_{i=1}^n|x_i|. \end{align*} To show that $$||x||_1 \leq \sqrt{n}||x||_2,$$ we let $x = (|x_1|, |x_2|, \cdots, |x_n|)$ and $y = (1,1, \cdots, 1)$. Then $$|x^Ty| \leq ||x||_2||x||_2 \Rightarrow |\sum_{i=1}^n|x_i|| \leq \sqrt{\sum_{i=1}^n|x_i|^2}\sqrt{\sum_{i=1}^n1},$$ but for each $i = 1,2, \ldots,n$, we have \begin{align*} |x_i| &>0,\\ |x_i|^2 &= x_i^2, \end{align*} hence it follows that $$|\sum_{i=1}^n|x_i|| = \sum_{i=1}^n|x_i| \text{ and } \sqrt{\sum_{i=1}^n|x_i|^2} = \sqrt{\sum_{i=1}^nx_i^2}.$$ Therefore, $$|\sum_{i=1}^n|x_i|| \leq \sqrt{\sum_{i=1}^n|x_i|^2}\sqrt{\sum_{i=1}^n1} \Rightarrow ||x||_1 \leq \sqrt{n}||x||_2,$$ as required. Is this approach correct?

Answer 1

Yes, this proof is correct. It's interesting that this inequality is actually not strict - consider $x = \left(1, 1, \dots, 1\right)^T$. And, btw, it's a particular case of the theorem stating all norms are equivalent in a finite-dimensional space.