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Prove that in $\mathbb R^n$, $d_1 (x,y)\le{\sqrt n}d_2 (x,y)$. Here $d_1(x,y)=\sum|x_i-y_i|$ and $d_2(x,y) = \sqrt{\sum (x_i-y_i)^2}$.

I tried to square both sides and expand but do not think that I can prove that $n\lvert x_i -y_i\rvert<\sum_{i=1}^n \lvert x_i -y_i\rvert$ for any i. Maybe this has something to do with Cauchy-Schwarz but I don't see how to use it.

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closed as unclear what you're asking by DeepSea, user91500, Harish Chandra Rajpoot, Tom-Tom, Clayton Sep 9 '15 at 14:37

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  • $\begingroup$ Please don't assume that every little notation your book uses is universally recognized by everyone. What exactly do you mean by $d_1, d_2$? $\endgroup$ – Paul Sinclair Sep 9 '15 at 5:19
  • $\begingroup$ $d_2$ is the euclidean metric $\endgroup$ – Ricc Sep 9 '15 at 5:20
  • $\begingroup$ Okay, the more obvious one is down. Now what about the other one? $\endgroup$ – Paul Sinclair Sep 9 '15 at 5:21
  • $\begingroup$ $d_1= \sum_{j=1}^n \lvert x_j - y_j\rvert$ $\endgroup$ – Ricc Sep 9 '15 at 5:22
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hint: $d_1 \leq \sqrt{n}\cdot d_2$ by CS inequality precisely.

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  • $\begingroup$ wow I suddenly see. So basically we let $b_n$ in CS inequality be 1, is that right? $\endgroup$ – Ricc Sep 9 '15 at 5:27
  • $\begingroup$ Yes, you got it. Good job ! $\endgroup$ – DeepSea Sep 9 '15 at 5:28
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Consider $f,g$ defined on the set $\{(x_1,x_2,\ldots, x_n)\in \mathbb{R}^n : x_i\geq 0 \quad\forall i\}$ by $$ f(x_1,x_2,\ldots, x_n)= \sum_{i=1}^n x_i, \text{ and } g(x_1,x_2,\ldots, x_n) = \sum_{i=1}^n x_i^2 $$ and we maximize $f$ under the constraint that $g = 1$. Lagrange multipliers will tell you that this is maximum when $$ x_i = \frac{1}{\sqrt{n}} \quad\forall 1\leq i\leq n \Rightarrow f(x_1,x_2,\ldots, x_n) = \sqrt{n} $$ Hence, for any $x\in \mathbb{R}^n$, consider $$ y = \frac{x}{d_2(x,y)} $$ Then $d_2(y,0) = 1$, whence $d_1(y,0) \leq \sqrt{n}$ and so $$ d_1(x,0) \leq \sqrt{n}d_2(x,0) $$ Now use the fact that $d_j(x,y) = d_j(x-y,0)$ for both $j=1, 2$.

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