jasperE3 wrote: Let $z_1,\ldots,z_n$ and $w_1,\ldots,w_n$ $(n\in\mathbb N)$ be complex numbers such that $$|\epsilon_1z_1+\ldots+\epsilon_nz_n|\le|\epsilon_1w_1+\ldots+\epsilon_nw_n|$$holds for every choice of $\epsilon_1,\ldots,\epsilon_n\in\{-1,1\}$. Prove that $$|z_1|^2+\ldots+|z_n|^2\le|w_1|^2+\ldots+|w_n|^2.$$ Since the inequality $$|\epsilon_1z_1+ \epsilon_2z_2 + \epsilon_3z_3\ldots+\epsilon_nz_n|\le|\epsilon_1w_1+ \epsilon_2z_2+\epsilon_3z_3\ldots+\epsilon_nw_n|$$ holds for all possible choices of $\epsilon_1,\ldots,\epsilon_n\in\{-1,1\}$, we have have $2^n$ number of inequalities corresponding to each choice of $\epsilon_1, \epsilon_2 \ldots,\epsilon_n$. Now squaring each of these $2^n$ inequalities and then adding all of them, we obtain $$ 2^n\left( |z_1|^2+|z_2|^2+\ldots+|z_n|^2 \right)\quad \quad \le \quad \quad 2^n\left(|w_1|^2+|w_2|^2+\ldots+|w_n|^2\right)$$ $$ \implies \quad \quad \quad |z_1|^2+\ldots+|z_n|^2\quad \quad\le \quad \quad |w_1|^2+\ldots+|w_n|^2$$