Consider the metric space $ \mathbb{R}^{n} $ with the usual metric.

Question

Consider the metric space $ \mathbb{R}^{n} $ with the usual metric. Prove that- $ \ \ max_{1 \leq k \leq n}|x_{k}-y_{k}| \leq d(X,Y) \leq \sqrt n \ \ max_{1\leq k \leq n} |x_{k}-y_{k}| $. $$ $$ Let $ x=(x_{1},x_{2},........,x_{n}) \ \ and \ \ y=(y_{1},y_{2},........,y_{n}) \in \mathbb{R}^{n} $. The usual metric is given by $ \ d(x,y)=||x-y||=[(x_{1}-y_{1})^{2}+..........+(x_{n}-y_{n})^{2}]^{1/2} $ . Then the square metric $ \rho(x,y)=Max_{1 \leq k \leq n} \{|x_{1}-y_{1}, ........, |x_{n}-y_{n}| \}$. Then we shall show that $ **\rho(x,y) \leq d(x,y) \leq \sqrt{n} \rho(x,y) $. **But how to show ? please help me

Answer 1

Let $m$ be a number between $1$ and $n$ such that $|x_m-y_m|$ is a maximum.

$$\begin{array}{rcl} d(X,Y) &=& \sqrt{(x_1-y_1)^2 + (x_2-y_2)^2 + \cdots + (x_n-y_n)^2} \\ &\ge& \sqrt{(x_m-y_m)^2 + 0 + \cdots + 0} \\ &=& |x_m-y_m| \\ &=& \displaystyle \max_{1 \le k \le n} |x_k - y_k| \\\\ d(X,Y) &=& \sqrt{(x_1-y_1)^2 + (x_2-y_2)^2 + \cdots + (x_n-y_n)^2} \\ &\le& \sqrt{(x_m-y_m)^2 + (x_m-y_m)^2 + \cdots + (x_m-y_m)^2} \\ &=& \sqrt{n(x_m-y_m)^2} \\ &=& \sqrt n |x_m-y_m| \\ &=& \sqrt n \displaystyle \max_{1 \le k \le n} |x_k - y_k| \end{array}$$