Dorin Marghidanu's Inequality in Complex Plane

Source

Dorin  Marghidanu's Inequality in Complex Plane, source

Problem

Dorin  Marghidanu's Inequality in Complex Plane, problem

Solution 1

Let's denote $S=z_1+z_2+\ldots+z_n.\,$ Then the required inequality becomes

(#)

$\displaystyle \sum_{k=1}^n|S-(n-1)z_k|\ge\sum_{k=1}^n|z_k|.$

Using the module (triangle) inequality, we have

$\displaystyle \begin{align} (n-1)\sum_{k=1}^n|z_k| &= \left|(n-1)S-(n-1)(S-z_k)\right|\\ &=\left|(n-1)S-(n-1)\sum_{\small{i=1,i\ne k}}^nz_i\right|\\ &=\left|\sum_{\small{i=1,i\ne k}}^n\left[S-(n-1)z_i\right]\right|\\ &\le \sum_{\small{i=1,i\ne k}}^n\left|S-(n-1)z_i\right|. \end{align}$

Therefore,

$\displaystyle \begin{align} (n-1)\sum_{k=1}^n\left|S-(n-1)z_k\right|&=\sum_{k=1}^n \sum_{\small{i=1,i\ne k}}^n\left|S-(n-1)z_i\right|\\ &\ge (n-1)\sum_{k=1}|z_k|, \end{align}$

which proves (#).

Equality occurs if $z_1=z_2=\ldots=z_n.$

Solution 2

Let

$\displaystyle w_i=\sum_{k}a_k z_k~\text{with}~a_k= \begin{cases} -(n-2),& \text{if } k=i \\ 1, & \text{otherwise}. \end{cases} $

Thus,

$\displaystyle \sum_{i=1,i\neq j}^n w_i = (n-1) z_j.$

From the triangle inequality,

$\displaystyle \sum_{i=1,i\neq j}^n |w_i| \geq (n-1) |z_j|.$

Thus,

$\displaystyle \begin{align} &\sum_{j=1}^n \sum_{i=1,i\neq j}^n |w_i| \geq (n-1)\sum_{j=1}^n |z_j|\\ &\Rightarrow (n-1) \sum_{k=1}^n |w_k| \geq (n-1)\sum_{j=1}^n |z_j| \\ &\Rightarrow LHS = \sum_{k=1}^n |w_k| \geq \sum_{j=1}^n |z_j| = RHS. \end{align}$

Acknowledgment

Dorin Marghidanu has kindly posted this problem at the CutTheKnotMath facebook page, along with a proof of his. Solution 2 is by Amit Itagi.

 

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