Let
$\eqalign{ & d\left( {x,y} \right) = \mathop {\max }\limits_{1 \leqslant i \leqslant n} \left\{ {\left| {{x_i} - {y_i}} \right|} \right\} \cr & d'\left( {x,y} \right) = \sqrt {\sum\limits_{i = 1}^n {{{\left( {{x_i} - {y_i}} \right)}^2}} } \cr & d''\left( {x,y} \right) = \sum\limits_{i = 1}^n {\left| {{x_i} - {y_i}} \right|} \cr} $
for any two points $x,y \in \Bbb R^n$.
How to prove the following holds?
$\eqalign{ & d\left( {x,y} \right) \leqslant d'\left( {x,y} \right) \leqslant \sqrt n \cdot d\left( {x,y} \right) \cr & d\left( {x,y} \right) \leqslant d''\left( {x,y} \right) \leqslant n \cdot d\left( {x,y} \right) \cr} $
I think I got the second one:
It is trivial that
$\mathop {\max }\limits_{1 \leqslant i \leqslant n} \left\{ {\left| {{x_i} - {y_i}} \right|} \right\} < \sum\limits_{i = 1}^n {\left| {{x_i} - {y_i}} \right|} $
Now let $k$ be the integer such that
$d\left( {x,y} \right) = \left| {{x_k} - {y_k}} \right|$
Then for each $1 \leq i \leq n$ we have that $\left| {{x_i} - {y_i}} \right|\leq \left| {{x_k} - {y_k}} \right|$
So summing from $1$ to $n$ one gets:
$d''\left( {x,y} \right) \leqslant n \cdot d\left( {x,y} \right)$