WLOG $|z_1|=a\geq |z_2|=b$ then $|z_1-z_2|+|z_1+z_2|\leq \sqrt{2|z_1-z_2|^2+2|z_1+z_2|^2}=\sqrt{4a^2+4b^2}\leq\sqrt{4a^2+4ab+b^2}=2a+b$

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.

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 $$AC+BD\leq a+(k+1)b$$ is always true.

@above, it's (almost) trivial