A problem regarding the in-equality of complex numbers . $\mathbf {The \ Problem \ is}:$ Let, $z_1,z_2 \cdots z_n$ be such that the real and imaginary parts of each $z_i$ are non-negative . Show that $$\bigg|\sum_{i=1}^n z_i\bigg| \geq \frac{1}{\sqrt2} \sum_{i=1}^n |z_i|.$$
$\mathbf {My \ approach} :$ Actually, at a very first glance, it seems that it can be proved using induction ,but I tried by using the triangular in-equality, though it didn't work .
I think there are some tricks related to it, a small hint is warmly appreciated .
 A: Hint: in conjunction with the two given expressions, also consider the expressions
$$
\bigg| \sum_{j=1}^n x_j + i \sum_{j=1}^n y_j \bigg|
\quad\text{and}\quad
\frac1{\sqrt2} \bigg( \sum_{j=1}^n x_j + \sum_{j=1}^n y_j \bigg).
$$
A: Since it was somewhere tagged as duplicate here my answer:
You may use 


*

*$\sqrt{a+b}\leq \sqrt a + \sqrt b$ for $a,b \geq 0$ and

*$(x+y)^2 \leq 2(x^2+y^2)$ (This is $2$-dimensional Cauchy-Schwarz inequality)


Let $z_k = x_k + iy_k$ for $k=1, \ldots , n$ where $x_k,y_k\geq 0$.
$$\left(\sum_{k=1}^n \lvert z_k \rvert\right)^2 =\left(\sum_{k=1}^n \sqrt{x_k^2 + y_k^2}\right)^2 \leq \left(\sum_{k=1}^n \left(x_k + y_k\right)\right)^2$$
$$= \left(\sum_{k=1}^n x_k + \sum_{k=1}^n  y_k\right)^2$$
$$\leq 2\left(\left(\sum_{k=1}^n x_k\right)^2 + \left(\sum_{k=1}^n  y_k\right)^2\right) = 2 \lvert \sum_{k=1}^n z_k \rvert^2$$
A: Let us define $\hat{m}=\frac{1}{\sqrt{2}}+j\ \frac{1}{\sqrt{2}}$ and $\hat{n}= -\frac{1}{\sqrt{2}}+j\ \frac{1}{\sqrt{2}}$ and $z_{i}=a_{i}\hat{m}+b_{i}\hat{n}$.
Since $z_{i}$ has non - negative real and imaginary part, 
$\left(b_{i}\right)^{2}\leq\left(a_{i}\right)^{2}\rightarrow\sum{\frac{1}{\sqrt{2}}\sqrt{\left(a_{i}^{2}+b_{i}^{2}\right)}}\leq\sum{a_{i}=\sqrt{\left(\sum{a_{i}}\right)^{2}}}\leq\sqrt{\left(\sum{a_{i}}\right)^{2}+ \left(\sum{b_{i}}\right)^{2}}$
aka
$\frac{1}{\sqrt{2}}\sum{\left|z_{i}\right|}\leq\left|\sum{z_{i}}\right|$
The $\sqrt{2}$ is a hint for me as it reminds me of $\sin{\left(\frac{\pi}{4}\right)}$ and $\cos{\left(\frac{\pi}{4}\right)}$. BTW could You share Your own answer too?
A: I'll provide a different approach to the ones mentioned so far.
First let $z_k=x_k+iy_k$, then
$$\vert z_1+\cdots+z_k\vert=\sqrt{(x_1+\cdots+x_k)^2+(y_1+\cdots+y_k)^2}$$
but $h(t):=\sqrt{t},\, t\in\mathbb R\,$ is a concave function, thus 
$$\sqrt{\frac{(x_1+\cdots+x_k)^2+(y_1+\cdots+y_k)^2}2}\stackrel{\color{red}{(*)}}{\geq} \frac{\sqrt{(x_1+\cdots+x_k)^2}+\sqrt{(y_1+\cdots+y_k)^2}}{2}$$
which implies that
\begin{align*}\sqrt{(x_1+\cdots+x_k)^2+(y_1+\cdots+y_k)^2}&\geq\frac{1}{\sqrt2}\left[(x_1+\cdots+x_k)+(y_1+\cdots+y_k)\right]\\
&=\frac{1}{\sqrt2}\left(\sqrt{(x_1+y_1)^2}+\cdots+\sqrt{(x_k+y_k)^2}\right)\\
&=\frac{1}{\sqrt2}\left(\sqrt{(x_1^2+2x_1y_1+y_1^2)}+\cdots+\sqrt{x_k^2+2x_ky_k+y_k^2}\right)
\end{align*}
If $\Re\left(z_k\right)>0$ and $\Im\left(z_k\right)>0$ for each $k$, then
$$\sqrt{x_k^2+2x_ky_k+y_k^2}\geq\sqrt{x_k^2+y_k^2}$$
so
\begin{align*}\sqrt{(x_1+\cdots+x_k)^2+(y_1+\cdots+y_k)^2}&\geq\frac{1}{\sqrt2}\left(\sqrt{x_1^2+y_1^2}+\cdots+\sqrt{x_k^2+y_k^2}\right)\\
&=\frac{1}{\sqrt2}\left(|z_1|+\cdots+|z_k|\right).
\end{align*}

As mentioned $\color{red}{(*)}$ follows from concavity: a real function is concave if and only if it is midpoint concave. Also, this is a particular case of Jensen's Inequality, specifically the finite form one, with $\varphi(x)=\sqrt{x}$.
