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):
$$|z + w|^2 + |z – w|^2 = 2|z|^2 + 2|w|^2$$
Best Answer
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.