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Compute distance in $l^3$ of element $\{\frac{1}{n+1} \}$ from set $\{ \{x_n\} l^3 : x_{100}=4x_{101} \}$

I've started:

$ dist= inf_{\{x_n\}}\{ (\sum_{n=0}|\frac{1}{n+1}-x_n|^3)^{\frac{1}{3}} \}= inf_{\{x_n\}}\{ (\sum_{n=0, n!=100, n!=101}|\frac{1}{n+1}-x_n|^3 + |\frac{1}{101}-x_{100}|^3|+ |\frac{1}{102}-x_{101}|^3|)^{\frac{1}{3}} \}= inf_{\{x_n\}}\{ (\sum_{n=0, n!=100, n!=101}|\frac{1}{n+1}-x_n|^3 + |\frac{1}{101}-4x_{101}|^3|+ |\frac{1}{102}-x_{101}|^3|)^{\frac{1}{3}} \} $

But here I've stuck. I don't know if I can assume, that for $ n!=100 $ and $ n!=101 $ $x_n=\frac{1}{n+1} $ to get those sum equals zero, and compute minimum of function $ |\frac{1}{101}-4x_{101}|^3|+ |\frac{1}{102}-x_{101}|^3$ ? Is my thought right? Or maybe I've just started from wrong side?

Thank you for any sugestion

pupilx
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    You approach is correct and indeed you have to choose ${x_n}=1/(n+1)$ for $n\neq 100,101$ to achieve an infimum, and hence minimize the function you obtained. – b00n heT Aug 28 '16 at 18:24
  • Thanks for your comment. Maybe it's already out of my question, but could you help me write the next step? I have problem with derivative of these function because of module – pupilx Aug 28 '16 at 18:50
  • An easy way to get rid of modules is to do a case division. In this case the three regions to be considered are $ x\geq \frac1{102}$, $\frac1{102}>x>\frac{1}{404}$, $x\leq \frac{1}{404}$. It's long, but easy and works every time – b00n heT Aug 28 '16 at 19:03
  • Ok, thanks, I've just thought that maybe there is any trick to solve it quicker – pupilx Aug 28 '16 at 19:06

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