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The statement of the problem: Let $n \in \mathbb N $ \ {0} and $z_1,z_2, ... z_n \in \mathbb C $. Prove that $$\sum_{i=1}^n |z_i||z-z_i| \ge \sum_{i=1}^n |z_i|^2$$ holds for any $z \in \mathbb C $ $\iff$ $z_1+z_2+...z_n=0$

My approach : I tried to prove the inequality using Cauchy's inequality and the modulus inequality, but I reached some ambiguous results, from which it does not appear that their sum would be 0. I think that the solution can also be a geometric one given the condition that z1 +z2+... zn = 0 it turns out that the vertices of the polygon with these affixes is regular, but I'm not sure.

Any and all proofs will be helpful. Thanks a lot!

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    $\begingroup$ Use \ge to get $\ge$ in MathJax $\endgroup$
    – jjagmath
    Feb 21 at 14:01
  • $\begingroup$ The inequality holds for $n=1$, $z=0$, and $z_1=1$, while $\sum_{i=1}^nz_1=1\neq0$. $\endgroup$ Feb 21 at 14:26
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    $\begingroup$ @JohnBentin: As I understand it, the task is to show that $$ \left( \forall z \in \Bbb C: \sum_{i=1}^n |z_i||z-z_i| \ge \sum_{i=1}^n |z_i|^2 \right) \iff z_1+z_2+\cdots + z_n=0 $$ $\endgroup$
    – Martin R
    Feb 21 at 14:30
  • $\begingroup$ OK, I get it. Put that way, I can't misread it. $\endgroup$ Feb 21 at 19:50

3 Answers 3

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Here's a proof using Triangle Inequality and Cauchy-Schwarz

(i.) when $\sum_{k=1}^n z_k = 0$
$\sum_{k=1}^n \vert z_k\vert^2= \Big \vert \sum_{k=1}^n \overline z_k \cdot z_k - \overline z_k\cdot z\Big\vert = \Big \vert \sum_{k=1}^n \overline z_k (z_k-z)\Big\vert$
$\leq \sum_{k=1}^n \Big \vert \overline z_k (z_k-z)\Big\vert=\sum_{k=1}^n \Big \vert z_k\Big \vert \cdot \Big \vert z-z_k\Big\vert$
by triangle inequality

(ii.) now suppose the inequality remains true when $\sum_{k=1}^n z_k =\lambda\neq 0$, then Cauchy-Schwarz tells us
$\sum_{k=1}^n \vert z_k\vert^2\leq \sum_{k=1}^n \Big \vert z_k\Big \vert \cdot \Big \vert z-z_k\Big\vert\leq \Big(\sum_{k=1}^n \vert z_k\vert^2\Big)^\frac{1}{2} \cdot \Big(\sum_{k=1}^n \Big \vert z-z_k\Big\vert^2\Big)^\frac{1}{2}$
$\implies\sum_{k=1}^n \vert z_k\vert^2\leq \sum_{k=1}^n \Big(z -\overline z_k\big)\overline{\Big(z - \overline z_k\Big)}= n \vert z\vert^2 + \Big(\sum_{k=1}^n \vert z_k\vert^2\Big)- 2\cdot \text{Re}\Big(z \sum_{k=1}^n z_k\Big)$
$\implies 2\cdot \text{Re}\Big(z\cdot \lambda\Big)\leq n \vert z\vert^2 $
which is obviously false, e.g. set $z:=\frac{\overline \lambda}{2n}$

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  • $\begingroup$ Thank you so much ! $\endgroup$
    – Last X
    Feb 22 at 7:57
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    $\begingroup$ @LastX I think the last part of the derivation is in error so that the last line should be $2\Re(\bar z\lambda)\le n|z|^2$. This has however no impact on the conclusion. $\endgroup$
    – user
    Feb 22 at 11:21
  • $\begingroup$ Since the inequality holds for all $z \in \mathbb C$ then $\sum_{k=1}^n \vert z_k\vert^2\leq \sum_{k=1}^n \Big \vert z-z_k\Big\vert^2\iff \sum_{k=1}^n \vert z_k\vert^2\leq \sum_{k=1}^n \Big \vert \overline z- z_k\Big\vert^2\iff \sum_{k=1}^n \vert z_k\vert^2\leq \sum_{k=1}^n \Big \vert z-\overline z_k\Big\vert^2$ so there is nothing inacurate in what I wrote. Sometimes the scaffolding used to get to the solution persists into the answer I post. $\endgroup$ Feb 22 at 17:20
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For convenience, we will use engineering notation $z^*$ for the complex conjugate of $z$ and statistical notation $\bar z$ for the arithmetic mean of the $z_i\;$ ($i=1,...,n$).

First, let us assume $$\sum_{i=1}^n |z_i||z-z_i|\geqslant \sum_{i=1}^n|z_i|^2 \quad\text{for all}\;z\in\Bbb C.\qquad(1)$$ We have $$\begin{align}0&\leqslant\sum_{i=1}^n(|z_i|-|z-z_i|)^2\\ &=\sum_{i=1}^n(|z_i|^2-2|z-z_i||z_i|+|z-z_i|^2)\\ &\leqslant \sum_{i=1}^n(|z_i|^2-2|z_i|^2+|z-z_i|^2)\quad\text{(by ineq. 1)}\qquad(*)\\ &=\sum_{i=1}^n(|z-z_i|^2-|z_i|^2)\\ &=\sum_{i=1}^n[(z-z_i)(z^*-z_i^*)-z_iz_i^*]\\ &=\sum_{i=1}^n(zz^*-z_iz^*-zz_i^*)\\ &=n(x^2+y^2)-2x\sum_{i=1}^nx_i-2y\sum_{i=1}^ny_i\quad\text{(where $z=x+\mathrm iy$ with $x,y\in\Bbb R$, etc.)}\\ &=n[(x-\bar x)^2+(y-\bar y)^2-\bar x^2-\bar y^2].\qquad(**) \end{align}$$ Overall, we may conclude that $0\leqslant (x-\bar x)^2+(y-\bar y)^2-\bar x^2-\bar y^2$ for all real $x$ and $y$. In particular, this is the case when $x=\frac12\bar x$ and $y=\frac12\bar y\,$: namely, $0\leqslant-\frac34(\bar x^2+\bar y^2)$. It follows that $\bar x=\bar y=0$.

To prove the converse implication, we will obtain a contradiction by assuming, contrary to inequality $1$, that there is $z_0=x_0+\mathrm iy_0$, with $x_0,y_0\in \Bbb R$, such that $$\sum_{i=1}^n|z_i||z_0-z_i|<\sum_{i=1}^n|z_i|^2\qquad(2)$$ while also $\sum_{i=1}^nz_i=0$. We use the same algebraic manipulations as above, but work backwards from line $**$ with $\bar x=\bar y=0$, replacing $z$ by $z_0$, and replacing $\leqslant$ by $>$ (from the contrary inequality $2)$ in line $*$. Consequently, we get

$$\sum_{i=1}^n(|z_i|-|z_0-z_i|)^2>\sum_{i=1}^n|z_0|^2.$$ Now write this as $\sum_{i=1}^n[(|z_i|-|z_0-z_i|)^2-|z_0|^2]>0$ and factorize to get $$\sum_{i=1}^n(|z_i|-|z_0-z_i|-|z_0|)(||z_i|-|z_0-z_i|+|z_0|)>0.$$ However, by the triangle inequality, the first factor of each term of the above sum is non-positive while the second factor is non-negative. So the sum is non-positive: the required contradiction.

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You could use the derivative mechanism with $z$ and $z^*$ (the complex conjugate) treated as independent variables. But if you aren't familiar with it, first study it to convince yourself that this is a consistent procedure for functions that are analytical in $z$ and in $z^*$, otherwise it would be black magic... We first define, for convenience: $$ X = \sum_{i=1}^n|z_i|\ |z-z_i| - \sum_{i=1}^n|z_i|^2 \geq 0 $$ We observe that we have $X=0$ in the origin ($z=0$). We now need to make the expression for $X$ analytical in $z$ and in $z^*$, like this: $$ X = \sum_{i=1}^n\left(|z_i|\ \sqrt{(z-z_i)(z^*-z_i^*)} - |z_i|^2\right) \geq 0 $$ Note that we do not care about the $z_i$, they can remain in the non-analytical |z_i|, we are only going to differentiate w.r.t. $z$ and $z^*$, giving: $$ \frac{dX}{dz^*} = \frac{d}{dz^*} \sum_{i=1}^n|z_i|\sqrt{(z-z_i)(z^*-z_i^*)} = \sum_{i=1}^n\frac{|z_i|\ (z-z_i)/2}{\sqrt{(z-z_i)(z^*-z_i^*)}} $$ $$ \mbox{ and similarly} \ \ \ \ \frac{dX}{dz} = \sum_{i=1}^n\frac{|z_i|\ (z^*-z_i^*)/2}{\sqrt{(z-z_i)(z^*-z_i^*)}}, $$ and then we look at the result in the origin where $z=z^*=0$, giving: $$ \left. \frac{dX}{dz^*}\right|_{z=0} = \sum_{i=1}^n\frac{|z_i|\ (-z_i)/2}{\sqrt{z_i z_i^*}} = -\frac{1}{2}\sum_{i=1}^n\ z_i, \ \ \ \ \mbox{and} \ \ \ \ \left. \frac{dX}{dz}\right|_{z=0} = -\frac{1}{2}\sum_{i=1}^n\ z_i^* $$ Now the finishing touch: because $X=0$ in the origin and has to be non-negative in any region around the origin it cannot have a finite first-order derivative! This means that the $\sum_{i=1}^n\ z_i$ that we encountered in the derivatives must be zero, $\frac{1}{2}$QED.

NB: without the originally given restriction, the expansion of $X$ around the origin would have been: $$ X = c + \alpha z + \alpha^* z^* + \beta z^2 + \beta^* (z^*)^2 + \gamma z z^* + \mbox{higher orders}, $$ but in this case we saw that $c=0$ and our given inequality then requires the first derivatives to be zero: $\alpha=\alpha^*=0$. With some further reasoning you can even find that $\beta$ should be $0$. We still might have $\gamma\neq 0$ (it should then be positive!)

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  • $\begingroup$ I haven't learned about derivatives yet, but thanks for the help! Do you know of any other solution to this problem that does not include derivatives ? $\endgroup$
    – Last X
    Feb 21 at 14:54
  • $\begingroup$ Not immediately... (but instead of those complex $z$ and $z^*$ derivatives you could rewrite the whole problem in terms of real 2-vectors with dot products and then use derivatives wrt. the real vector components). $\endgroup$ Feb 21 at 15:11
  • $\begingroup$ what means z* ? $\endgroup$
    – Last X
    Feb 21 at 15:21
  • $\begingroup$ Complex conjugate! (And for the derivative wrt. $z^*$, if you want to study it look on the web, e.g. here: physicsforums.com/threads/… If you want to go to 2-vectors instead, the derivative will just be wrt. the x and y-components so then it doesn't matter. But it is clumsy to first have to translate to vectors and then doing every derivative for two variables instead of one.) $\endgroup$ Feb 21 at 15:33
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    $\begingroup$ Why did you replace the inequality with equality? $\endgroup$
    – user
    Feb 21 at 16:05

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