# min $k$ s.t. $|z_1+z_2+\cdots+z_n|\geq \frac{1}{k}(|z_1|+|z_2|+\cdots+|z_n|).$

Find the smallest positive real number $$k$$ such that, given any finite set $$z_1,\cdots, z_n$$ of complex numbers, all with strictly positive real and imaginary parts, the following inequality holds: $$|z_1+z_2+\cdots+z_n|\geq \frac{1}{k}(|z_1|+|z_2|+\cdots+|z_n|).$$

Answer- $$\sqrt{2}$$

My Attempt:

First, we take $$n=2$$. Let $$z_i=r_ie^{i\theta_i}$$ for $$i=1, 2$$. Then $$|z_1+z_2|^2=|r_1e^{i\theta_1}+r_2e^{i\theta_2}|^2= r_1^2+r_2^2+r_1r_2e^{i(\theta_1-\theta_2)}+r_1r_2e^{i(\theta_2-\theta_1)}.$$ Also $$|z_1|+|z_2|=r_1+r_2.$$ Therefore, the given inequality holds if
$$r_1^2+r_2^2+r_1r_2e^{i(\theta_1-\theta_2)}+r_1r_2e^{i(\theta_2-\theta_1)}\geq \frac{1}{k^2}(r_1+r_2)^2$$ $$\implies (k^2-1)(r_1^2+r_2^2)+r_1r_2(k^2 e^{i(\theta_1-\theta_2)}+k^2e^{i(\theta_2-\theta_1)}-2)\geq 0.$$ which holds if $$k^2(e^{i(\theta_1-\theta_2)}+e^{i(\theta_2-\theta_1)})\geq 2$$

• It is not difficult to show that the inequality holds with $k=\sqrt 2$. But I doubt that this is the smallest value if $n \ge 3$. Sep 4, 2021 at 10:09
• @MartinR Actually it holds for $k=\sqrt{2}$. Sep 4, 2021 at 11:07
• @YiorgosS.Smyrlis: Yes, that is what I said. It is also easy to see that this is the best possible constant in the case of two numbers. My misunderstanding was that I thought that the question asks for the best constant for a fixed number $n$ (which looks more difficult to me if $n \ge 3$). Sep 4, 2021 at 11:35
• This should also directly follow from minkowski or holder if you rewrite the question in terms of its positive parts. Sep 15, 2021 at 14:22

Set $$w_k=e^{-i\pi/4}z_k$$. Then $$w_k=|z_k|e^{i\theta_k}$$, where $$\theta_k\in (-\pi/4,\pi/4)$$, and $$\cos\theta_k>1/\sqrt{2}$$. Thus $$\mathrm{Re}\,w_k>\frac{1}{\sqrt{2}}|w_k|.$$ Then $$\left|\sum_{k=1}^n z_k\right|=\left|\sum_{k=1}^n w_k\right|\ge\sum_{k=1}^n \mathrm{Re} \,w_k =\sum_{k=1}^n |w_k|\cos\theta_k >\frac{1}{\sqrt{2}}\sum_{k=1}^n |w_k|=\frac{1}{\sqrt{2}}\sum_{k=1}^n |z_k|.$$
$$k=\sqrt{2}$$ is the smallest.
If $$z_1=1+i\varepsilon$$ and $$z_2=\varepsilon+i$$, $$\varepsilon>0$$, then
$$|z_1+z_2|=(1+\varepsilon)\sqrt{2}, \qquad |z_1|+|z_2|=2\sqrt{1+\varepsilon^2}, \quad \frac{|z_1+z_2|}{|z_1|+|z_2|}=\frac{1}{\sqrt{2}}\cdot\frac{1+\varepsilon}{\sqrt{1+\varepsilon^2}}$$ and $$\inf_{\varepsilon>0}\frac{|z_1+z_2|}{|z_1|+|z_2|}=\frac{1}{\sqrt{2}}.$$
• I think what you tried to mean is the following $:$ \begin{align*} \left |\sum\limits_{k=1}^{n} z_k \right | & = \left |\sum\limits_{k=1}^{n} w_k \right | \\ & \geq \text {Re} \left ( \sum\limits_{k=1}^{n} w_k \right ) \\ & = \sum\limits_{k=1}^{n} \text {Re}\ (w_k) \\ & \gt \frac {1} {\sqrt {2}} \sum\limits_{k=1}^{n} |w_k| \\ & = \frac {1} {\sqrt {2}} \sum\limits_{k=1}^{n} |z_k|. \end{align*} Done! BTW very nice answer.+1