Showing that $\sqrt{\sum_{k=1}^{n} |u_k|^2}\leq \sum_{k=1}^{n}|u_k| \leq n\cdot\max\{|u_k|:1\leq k \leq n\}$

Question

My textbook has used the following inequalities without any proof:

Let $X=\Bbb R^n$, then

$$\sqrt{\sum_{k=1}^{n} |u_k|^2}\leq \sum_{k=1}^{n}|u_k| \leq n\cdot \max\{|u_k|:1\leq k \leq n\} \leq n\cdot\sqrt{\sum_{k=1}^{n}|u_k|^2}.$$

The second inequality is obvious. But I am struggling with others.

Answer 1

$$\left(\sum_{k=1}^{n}|u_k|\right)^2{=\sum_{k,l=1}^{n}|u_k|\cdot|u_l|\\=\sum_{k=1}^{n}|u_k|^2+\sum_{k,l=1\\k\ne l}^{n}|u_k|\cdot |u_l|\\\ge \sum_{k=1}^{n}|u_k|^2}$$therefore $$\sqrt {\sum_{k=1}^{n}|u_k|^2}\le \sum_{k=1}^{n}|u_k|$$

Answer 2

The right-most inequality holds since $$|u_k|\leq\max\{|u_i|, 1\leq i\leq n\}$$ for every $k$ so that $$\sum_{k=1}^n|u_k|\leq\sum_{k=1}^n \max\{|u_i|\}=n\max\{|u_i\}$$ since we sum a constant.

To prove the left-most inequality, note that it is enough to prove $$\sum_{k=1}^n |u_k|^2 \leq \left(\sum_{k=1}^n |u_k|\right)^2$$ which should be fairly obvious if you expand the right-hand side.