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 $i$?

  • 2
    $\begingroup$ I think you just take $j$ is the index such that $|x_j-z_j|=\max\{|x_i-z_i|\}$ and now $|x_j-z_j|\le|x_j-y_j|+|y_j-z_j|\le\max\{|x_i-y_i|\}+\max\{|y_i-z_i|\}$? $\endgroup$
    – Frank Kong
    Nov 6, 2021 at 19:21

1 Answer 1


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.

  • $\begingroup$ How do you take max of both sides in your last line? This needs a proof. RHS is const and LHS variable. Taking max of LHS may make it greater then LHS $\endgroup$ Nov 6, 2021 at 19:34
  • $\begingroup$ 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 $i \in \{1,\ldots,n\}$ $\endgroup$
    – Sebathon
    Nov 6, 2021 at 19:50
  • $\begingroup$ I missed the "For each i" part. Then its a deal :') $\endgroup$ Nov 6, 2021 at 20:11

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