Equivalent metrics - proof

Question

Why are $d_1$ and $d_2$ equivalent?

$$d_1(x,y)=\sqrt{\sum_{i=1}^{n} (x_i-y_i)^2}$$ $$d_2(x,y)=\max_i|x_i-y_i|$$

I'm stuck, I started here:

$$d_1(x,y)=\sqrt{\sum_{i=1}^{n} (x_i-y_i)^2}=\sum_{i=1}^{n} |x_i-y_i|$$

Answer 1

One has the following inequality: $$d_1(x,y)=\sqrt{\sum_{i=1}^{n} (x_i-y_i)^2} \leq \sqrt{\sum_{i=1}^{n} \max_i \vert x_i-y_i \vert^2} = \sqrt n \cdot \max_i \vert x_i-y_i \vert = \sqrt n \cdot d_2(x,y).$$ Further there exists a $j \in \{1, \dots, n\}$ such that $\vert x_j - y_j \vert = \max_i \vert x_i-y_i \vert$. Thus we can deduce $$d_2(x,y)=\max_i \vert x_i-y_i \vert = \vert x_j - y_j \vert = \sqrt{( x_j - y_j )^2} \leq \sqrt{\sum_{i=1}^{n} (x_i-y_i)^2} = d_1(x,y).$$ Hence the two metrics are equivalent. Futher notice that the square root is no linear function. This is why your ansatz fails.

Answer 2

Hint:

Consider that: $$\max_{i=1,\dots,n}|x_i-y_i|=\max_{i=1,\dots,n}\sqrt{(x_i-y_i)^2}\leq\sqrt{\sum_{i=1}^n(x_i-y_i)^2}\leq\sqrt{\sum_{i=1}^n\max_{i=1,\dots,n}(x_i-y_i)^2}\\ =\sqrt{n\max_{i=1,\dots,n}(x_i-y_i)^2}=\sqrt{n}\max_{i=1,\dots,n}|x_i-y_i|$$