Let z1 and z2 be complex numbers. Prove that |z1+z2|+|z1−z2|⩽Vlad Cerbu and Sorin Rădulescu
Problem
Source: Romania National Olympiad 2022
Tags: complex numbers, romania, inequalities
21.04.2022 09:36
WLOG |z_1|=a\geq |z_2|=b then |z_1-z_2|+|z_1+z_2|\leq \sqrt{2|z_1-z_2|^2+2|z_1+z_2|^2}=\sqrt{4a^2+4b^2}\leq\sqrt{4a^2+4ab+b^2}=2a+b
21.04.2022 11:15
Also possible by geometric interpretation By plotting the points on the complex plane, the problem reduces to this: (deal with degenerate cases apart) Lemma Let ABCD be a parallelogram with AB\le AD. Then AC+BD \le 2AD+AB. Proof sketch By smoothing by alterating distance AD by moving D without changing line AD, you can assume WLOG that AD=AB, thus ABCD is a rhombus. To finish, consider the reflection A' of A across BC, and look at the rectangle triangle ACA'. Just need to prove that AC+A'C \le AB+BC+BA'. To finish, just use AM-QM and pythagoran theorem: AC+A'C \le 2\sqrt{\frac{AC^2+A'C^2}{2}} = 2\sqrt{2}AB < 3AB = AB+BC+BA' and the result follows. I guess equality holds for some degenerate case.
25.04.2022 18:36
oVlad wrote: Let z_1 and z_2 be complex numbers. Prove that |z_1+z_2|+|z_1-z_2|\leqslant |z_1|+|z_2|+\max\{|z_1|,|z_2|\}.Vlad Cerbu and Sorin Rădulescu See here If ABCD is an arbitrary non-degenerate parallelogram with AB:=a\leq b:=AD, then find the real numbers k for which AC+BD\leq a+(k+1)bis always true.
27.04.2022 06:35
@above, it's (almost) trivial