Vector norm and relationship with euclidean distance If $y\in E_n$ (n dimensional euclidean space) show that $||\textbf{y}||\leq|\textbf{y}|\leq \sqrt{n}||\textbf{y}||$ 
Where $||\textbf{y}||$ is the euclidean length of the vector $\textbf{y}$
and $|\textbf{y}|$ is the norm of $\textbf{y}$
The hint: Use the inequality $2|uv|\leq |u|^2+|v|^2$ 
to show 
$||\textbf{y}||^2\leq|\textbf{y}|^2\leq n||\textbf{y}||^2$
By letting $u=|y|$ and $v=y$ I get
$2||y|y|\leq||y||^2+|y|^2$ I don't really see how this helps.


Let $y=(y_1,...,y_n).$ $$\text { We have }\quad |y|^2-\|y\|^2=[\;\sum_{i=1}^n|y_i|\;]^2-\sum_{i=1}^n y_i^2=\sum_{1\leq i<j\leq n}2|y_i|.|y_j|\geq 0.$$  $$\text  { We have }\quad 0\leq \sum_{1\leq i<j\leq n}(y_i-y_j)^2=(n-1)\sum_{i-1}^ny_i^2-2\sum_{1\leq i<j\leq n}y_iy_j=$$ $$=n\|y\|^2-\sum_{i=1}^ny_i^2-2\sum_{1\leq i<j\leq n}y_iy_j=n\|y\|^2-|y|^2.$$
