# If $|z_1\pm z_2\pm\ldots\pm z_n|\le|w_1\pm w_2\pm\ldots\pm w_n|$, prove that $|z_1|^2+|z_2|^2+\ldots+|z_n|^2\le|w_1|^2+|w_2|^2+\ldots+|w_n|^2$.

Let $z_1, z_2, \ldots, z_n$ and $w_1, w_2,\ldots,w_n$ be some complex numbers, and $n$ a positive integer greater than $1$. If $$|z_1 \pm z_2 \pm \ldots \pm z_n| \le |w_1 \pm w_2 \pm \ldots \pm w_n|$$ Prove $$|z_1|^2 + |z_2|^2 + \ldots + |z_n|^2 \le |w_1|^2 + |w_2|^2 + \ldots + |w_n|^2$$

Let $z_k = a_k + ib_k$ and $w_k = c_k + id_k$. The two relations become: $$\left(a_1 + a_2 + \ldots + a_n\right)^2 + \left(b_1 + b_2 + \ldots + b_n\right)^2 \le \left(c_1 + c_2 + \ldots + c_n\right)^2 + \left(d_1 + d_2 + \ldots + d_n\right)^2$$ $$a_1^2 + b_1^2 + a_2^2 + b_2^2 + \ldots + a_n^2 + b_n^2 \le c_1^2 + d_1^2 + c_2^2 + d_2^2 + \ldots + c_n^2 + d_n^2$$

Can you give me a hint, please? Thanks!

Maybe this helps you (it's the first part of the problem): $$|z + w|^2 + |z - w|^2 = 2|z|^2 + 2|w|^2$$

• Induction should be better in general. – Nosrati Jul 23 '18 at 11:40
• @mrtaurho Your inequality is false: take $n=2,z_1=1,z_2=i$ – Kavi Rama Murthy Jul 23 '18 at 11:55
• Yeah, I just realized for myself. I am sorry for the wrong hint. – mrtaurho Jul 23 '18 at 11:58
• Hint: $$\sum\limits_{\epsilon_2,\cdots,\epsilon_n \in \{-1,1\}}|z_1 + \epsilon_2 z_2 + \cdots + \epsilon_n z_n|^2 = 2^{n-1}\left(|z_1|^2 + \cdots + |z_n|^2\right)$$ – achille hui Jul 23 '18 at 12:00

For $n=2$ we have $|z_1+z_2|^{2}+|z_1-z_2|^{2}=2|z_1|^{2}|+2|z_2|^{2}$. Using a similar formula for the $w$'s we get the result immediately. The above formula for $n=2$ generalizes to any $n$ and (with the constant $2$ on the right side replaced by $2^{n}$) and so the same proof works for any $n$. In all cases what you do is write down the inequality for all possible plus minus signs and add them up.
• The constant on the right side is replaced by $2^n$, not by $n$. – uniquesolution Jul 23 '18 at 11:57
• the factor in RHS is $2^{n-1}$, not $2^n$. – achille hui Jul 23 '18 at 12:03
• @achillehui - You are wrong. The generalized parallelogram identity says that the average of $\sum_{i=1}^n\|\pm v_i\|^2$ over all possible signs $\pm 1$ is exactly $\sum_i\|v_i\|^2$. There are $2^n$ possible choices of signs in $n$ dimensions. – uniquesolution Jul 23 '18 at 12:07
• @uniquesolution The first $z_1$ doesn't have a -ve sign attached to it. – achille hui Jul 23 '18 at 12:09