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Let $ Z_1 , Z_2 ,.... , Z_{2n} $ be a distinct collection of complex numbers such that $$ | Z_{2i} - Z_{2i - 1} | \geq \max_{1\leq j, k \leq 2i} | Z_j - Z_k | \quad \forall i, 1\leq i \leq n$$ Define $ c = \frac{ Z_1 + Z_2}{2} $ . $$\\$$ Show that, for any $ i , 1\leq i \leq n$ , $$ | Z_{2i} - Z_{2i - 1} | \geq | Z_{2i} - c| $$

I figured out that any sequence of distinct complex numbers can be rearranged in such a manner, and also a list of inequalites, which are

$$ | Z_{2i} - Z_{2i - 1} | \geq | Z_{2i} - Z_1 | $$ $$ | Z_{2i} - Z_{2i - 1} | \geq | Z_{2i-1} - Z_1 | $$ $$ | Z_{2i} - Z_{2i - 1} | \geq | Z_2 - Z_{2i} | $$ $$ | Z_{2i} - Z_{2i - 1} | \geq | Z_2 - Z_{2i-1} | $$

$$ | Z_{2i} - Z_{2i - 1} | \geq | Z_2 - Z_1 | = | Z_2 - c | + | c - Z_1 | $$

I need to somehow use triangle inequality, and the above inequalities, to get the above result. Any hints would be welcome .

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See if this can give you any help:

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Since

$|Z_{2i}-Z_{2i-1}|\geq|Z_{2i}-Z_2|$

$|Z_{2i}-Z_{2i-1}|\geq|Z_{2i}-Z_1|$

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we have that

$\begin{align*} 2|Z_{2i}-Z_{2i-1}|&\geq|Z_{2i}-Z_2|+|Z_{2i}-Z_1|\\ &\geq|(Z_{2i}-Z_2)+(Z_{2i}-Z_1)|\qquad\quad(\text{triangle inequality})\\ &=|2Z_{2i}-(Z_1+Z_2)|\\ \end{align*}$

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then multiply $\frac12$ at both sides, we get

$\begin{align*} |Z_{2i}-Z_{2i-1}|&\geq|\frac{2Z_{2i}-(Z_1+Z_2)}2|\\ &=|Z_{2i}-\frac{Z_1+Z_2}2|\\ &=|Z_{2i}-c|\\ \end{align*} $

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