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 $$k^2(e^{i(\theta_1-\theta_2)}+e^{i(\theta_2-\theta_1)})\geq 2$$
I struck at this point. Please help.
Best Answer
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 $\cos\theta_k\ge 1/\sqrt{2}$. Thus $$ \mathrm{Re}\,w_k\ge\frac{1}{\sqrt{2}}|w_k|. $$ Then $$ \left|\sum_{k=1}^n z_k\right|=\left|\sum_{k=1}^n w_k\right|\ge\sum_{k=1}^n \mathrm{Re} \,w_k =\sum_{k=1}^n |w_k|\cos\theta_k \ge\frac{1}{\sqrt{2}}\sum_{k=1}^n |w_k|=\frac{1}{\sqrt{2}}\sum_{k=1}^n |z_k|. $$
Also, $k=\sqrt{2}$ 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 $$ \inf_{\varepsilon>0}\frac{|z_1+z_2|}{|z_1|+|z_2|}=\frac{1}{\sqrt{2}}. $$