Let $d$ be the distance function defined on $\mathbb{R}^n$ by the correspondence $d(x,y) = \max_{1\leq i \leq n} \{d_i(x_i,y_i)\}$ for $x = (x_1,x_2,...,x_n),y=(y_1,y_2,...,y_n)\in \mathbb{R}^n,$ and let $d'$ be the euclidean distance function $d'(x,y) = \sqrt{\sum_{i=1}^n(x_i-y_i)^2}$. Then, for each pair of points $x,y \in \mathbb{R}^n$, $$d(x,y) \leq d'(x,y) \leq \sqrt{n} * d(x,y).$$ That is, $$\max_{1\leq i \leq n} \{d_i(x_i,y_i)\} \leq \sqrt{\sum_{i=1}^n(x_i-y_i)^2} \leq \sqrt{n}*\max_{1\leq i \leq n} \{d_i(x_i,y_i)\}.$$
How can I show the second part of this inequality? Namely that $\sqrt{\sum_{i=1}^n(x_i-y_i)^2} \leq \sqrt{n}*\max_{1\leq i \leq n} \{d_i(x_i,y_i)\}$.