$|z_1+z_2+…+z_n|\ge \frac{1}{\sqrt{1+t^2}}(|z_1|+|z_2|+…|z_n|)$

Asked Jul 06 '22 at 12:31

Active Jul 06 '22 at 15:52

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Let $z_1,z_2,…,z_n\in \mathbb C$, such that $|y_i|\le tx_i$ for some positive number $t$ ($z_i=x_i+iy_i$). I want to verify the inequality $|z_1+z_2+…+z_n|\ge \frac{1}{\sqrt{1+t^2}}(|z_1|+|z_2|+…|z_n|)$.

I found that $\frac{1}{\sqrt{1+t^2}}(|z_1|+|z_2|+…|z_n|)\le (x_1+x_2+…+x_n)$ while $|z_1+z_2+…+z_n|\ge (1-t) (x_1+x_2+…+x_n)$. But this doesn’t imply the inequality. Any hint?

Update: I found that this inequality has a simple geometric interpretation. For the “polygonal line” $z_1,z_2,…,z_n$ whose sides $z_i$ obey the angle condition $|\frac{y_i}{x_i}|\le t = \tan \alpha$, the length of the polygonal line (or precisely, the length of the resultant) is greater than or equal to the sum of the lengths of their corresponding projections onto the $x$-axis.

edited Jul 06 '22 at 15:52

asked Jul 06 '22 at 12:31

Why did my post get a downvote? – Jul 06 '22 at 12:37
3

You almost solved it. Just notice that the factor $(1-t)$ you found can be a $1$ – Exodd Jul 06 '22 at 12:40
@Exodd $t$ is a fixed positive number. How could the factor be $1$? – Jul 06 '22 at 12:45
I mean, if you substitute $1$ to your $(1-t)$ the inequality still holds – Exodd Jul 06 '22 at 12:46
@Exodd right, it’s just $|x|\le |z|$. – Jul 06 '22 at 12:50

$|z_1+z_2+…+z_n|\ge \frac{1}{\sqrt{1+t^2}}(|z_1|+|z_2|+…|z_n|)$

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