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Suppose $d$ is the Euclidean metric on $\mathbb{R}^n$, $d(x,y)=\sqrt{\sum_{i=1}^n(x_i-y_i)^2}$, and $\tau(x,y)=\sum_{i=1}^n|x_i-y_i|$ is the taxicab metric on $\mathbb{R}^n$.

When showing this two metrics are equivalent, it is not hard to bound $\tau$ by $$ \tau(x,y)\leq n d(x,y). $$

Is it possible to tighten the bound to $\tau(x,y)\leq\sqrt{n}d(x,y)$?

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No, it's not possible: $\tau( (0.4,0.4), (0,0) ) = 0.8 > 0.46 > \sqrt{2} (0.4^2+0.4^2) = \sqrt{2}d( (0.4,0.4), (0,0) )$

edit: As I somehow missed to take the square root, this counterexample is wrong. See below for a correct answer that it is in fact possible to work with $\sqrt{n}$.

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  • $\begingroup$ Thanks for counter-example. $\endgroup$ – Jong Gim May 24 '13 at 1:17
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I think the counterexample forgot to take the square root of $d$. Instead, this follows from the Cauchy-Schwarz inequality.

If $x,y$ are in $\mathbb{R}^n$, $$ \tau(x,y)=\sum_{i=1}^n|x_i-y_i|=\left|\sum_{i=1}^n|x_i-y_i|\right|\leq\sqrt{\sum_{i=1}^n (x_i-y_i)^2}\sqrt{\sum_{i=1}^n 1^2}=\sqrt{n}d(x,y). $$

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  • $\begingroup$ Oh, you are absolutly right... I somehow forgot to take the square root - weird... $\endgroup$ – user79202 May 24 '13 at 20:03
  • $\begingroup$ @user79202 Happens to all of us at some point! $\endgroup$ – Ben West May 24 '13 at 22:54

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