# For $x,y,z \in \mathbb{R}^n$ is it true that $\max\{|x_i-z_i|\} \leq \max\{|x_i-y_i|\} + \max\{|y_i-z_i|\}$ for $1 \leq i \leq n$?

For $$x,y,z \in \mathbb{R}^n$$ is it true that $$\max\{|x_i-z_i|\} \leq \max\{|x_i-y_i|\} + \max\{|y_i-z_i|\}$$ for $$1 \leq i \leq n$$?

Here I want to use the metric $$d(x,y)=|x_1-y_1|+...+|x_n-y_n|$$ to prove that the Triangle inequality holds for $$d_{max}=\max\{|x_i-y_i|: 1 \leq i \leq n\}$$.

Since $$d$$ is a metric on $$\mathbb{R}^n$$ we have that for any $$x,y,z \in \mathbb{R}^n$$, $$d(x,z)\leq d(x,y)+d(y,z)$$. Then for $$1 \leq i \leq n$$ we have, $$|x_i-z_i| \leq |x_i-y_i|+|y_i-z_i|$$

I am having trouble explicitly linking this statement to the desired result. In other words, precisely why does this inequality still hold when we take the maximum of the sets over $$i$$?

• I think you just take $j$ is the index such that $|x_j-z_j|=\max\{|x_i-z_i|\}$ and now $|x_j-z_j|\le|x_j-y_j|+|y_j-z_j|\le\max\{|x_i-y_i|\}+\max\{|y_i-z_i|\}$? Nov 6, 2021 at 19:21

For each $$i$$ thanks to the triangular inequality you have $$|x_{i}-z_{i}| \leq |x_{i}-y_{i}|+|y_{i}-z_{i}|.$$ Now, note that $$|x_{i}-y_{i}| \leq \max \{ |x_{i}-y_{i}| \}$$, and $$|y_{i}-z_{i}| \leq \max \{ |y_{i}-z_{i}| \}$$. Hence, $$|x_{i}-z_{i}| \leq \max \{ |x_{i}-y_{i}| \} +\max \{ |y_{i}-z_{i}| \}.$$ Now take the maximum in this expression to obtain the desire inequality.
• The RHS is constant. Is like taking the maximum to the inequality $$|x_{i}-z_{i}|\leq a+b.$$ Is because the inequality is true for every $i \in \{1,\ldots,n\}$ Nov 6, 2021 at 19:50