How can I prove $d_1(x,y) \leq n d_\infty (x,y)$ $$d_\infty (x,y) = \max{|x_i - y_i| | i=1,2,...,n}$$
$$d_1 (x,y)= \sum_{i=1}^n |x_i - y_i|$$
$$d_\infty : \mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}$$
$$d_1: \mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}$$
$$x,y\in \mathbb{R}^n$$
I would like to prove that $d_1(x,y) \leq n d_\infty (x,y)$.
Attempt:
Since $$d_1(x,y) = \sum_{i=1}^n |x_i - y_i|=|x_1 - y_1|+ \dots + |x_n - y_n|\leq \max|x_1 - y_1| +\dots + \max|x_i - y_i |=$$ $$= n\max |x_i - y_i |=nd_{\infty}(x,y).$$ 

As C. Falcon says, you should change $d_{\infty}$ by $d_{2}$.
By Cauchy-Schwarz we have $$(\sum_{i=1}^n (|x_i - y_i|\cdot 1))^2\le (\sum_{i=1}^n |x_i - y_i|^2)(1^2+\cdots 1^2).$$ 
Then $$\sum_{i=1}^n |x_i - y_i|\le \sqrt{n}(\sqrt{\sum_{i=1}^n |x_i - y_i|^2}).$$ 
Hence $d_{1}(x,y)\le \sqrt{n}d_{2}(x,y)$.
