Clarification needed for a question about norms

Question

I have to show that for all $x \in \mathbb{R}^{n}$ the following inequality holds: $$ \sqrt{n} \|x\|_2 \le n \|x\|_\infty.$$ I also have the following answer: $$\|x\|_2= \left( \sum_{i=1}^n |x_i|^2 \right)^{1/2} \le \sqrt{n} \max_{1\le i \le n} |x_i|= \sqrt{n} \|x\|_\infty$$

Note that I understand that $\|x\|_2 \le \sqrt{n} \|x\|_\infty$ implies the required proof is correct, since we can multiple both sides by $\sqrt{n}$ to get it in the required form.

However, I cannot seem to grasp the rest. Any help will be appreciated.

Answer 1

$\|x\|_\infty=\sup(|x_i|)$. This implies that

$\|x\|_2=\sqrt{\sum_{i=1}^{i=n} x_i^2}\leq \sqrt{\sum_{i=1}^{i=n}\|x\|^2_\infty}=\sqrt{n}\|x\|_\infty$.

Answer 2

$ |x_i^{2}| \leq \|x\|_{\infty}^{2}$ for each $i$. Add these up and take square root.

Answer 3

Thank you for all your comments, Tsemo made an error placing the sum outside the square root, but with the replies of Kavi and Tsemo I managed to write the following result:

$$\|x\|_2=\left( \sum_{i=1}^n |x_i|^2 \right)^{1/2} \le \left( \sum_{i=1}^n \left( \max_{1\le i \le n} |x_i| \right)^2 \right)^{1/2}= \sqrt{n} \max_{1\le i \le n} |x_i|$$

Thank you very much!