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$\mathbf {The \ Problem \ is}:$ 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 .

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  • $\begingroup$ A straightforward approach should work. Establish that $|z_1 + z_2| \geq \frac{|z_1|+|z_2|}{\sqrt2}$. You can then use induction to obtain the result. $\endgroup$ – Brian Tung Mar 4 at 7:02
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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 $x_k,y_k\geq 0$.

$$\left(\sum_{k=1}^n \lvert z_k \rvert\right)^2 =\left(\sum_{k=1}^n \sqrt{x_k^2 + y_k^2}\right)^2 \leq \left(\sum_{k=1}^n \left(x_k + y_k\right)\right)^2$$ $$= \left(\sum_{k=1}^n x_k + \sum_{k=1}^n y_k\right)^2$$ $$\leq 2\left(\left(\sum_{k=1}^n x_k\right)^2 + \left(\sum_{k=1}^n y_k\right)^2\right) = 2 \lvert \sum_{k=1}^n z_k \rvert^2$$

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Hint: in conjunction with the two given expressions, also consider the expressions $$ \bigg| \sum_{j=1}^n x_j + i \sum_{j=1}^n y_j \bigg| \quad\text{and}\quad \frac1{\sqrt2} \bigg( \sum_{j=1}^n x_j + \sum_{j=1}^n y_j \bigg). $$

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  • $\begingroup$ great hint: I couldn't solve the original task, but was able to solve it given your answer. but wondering how to come up to the hint itself $\endgroup$ – govordovsky Mar 4 at 7:13
  • $\begingroup$ I found one part of the hint by realizing that if the $n$ numbers $z_1,\dots,z_n$ were replaced by the $2n$ numbers $x_1,\dots,x_n,iy_i,\dots,iy_n$, then the left-hand side wouldn't change but the right-hand side would increase. Thus we might as well consider this case (where all numbers are either real or imaginary), since it really implies the full case. The other part of the hint came from trying to understand where the $\sqrt2$ might come from, by thinking about the case $n=1$ (which is trivial in the original formulation but slightly nontrivial in the new one). $\endgroup$ – Greg Martin Mar 4 at 17:31
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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?

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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 $\varphi(x)=\sqrt{x}$.

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